circle of problem solving

Area of Circle Exercises

Area of a circle practice problems with answers.

There are twelve (12) practice problems in this exercise about the area of the circle . You may use a calculator. Do not round intermediate calculations.

Round your final answer to two decimal places unless the exact answer is required.

For your convenience, I have included the different variations of formulas that you can use to find the area of a circle.

Problem 1: What is the area of a circle with radius [latex]8[/latex] meters? Leave your answer in terms of [latex]\large{\pi}[/latex].

This problem requires us to leave our answer in terms of [latex]\pi[/latex].

[latex]64\pi [/latex] square meters

Problem 2: The diameter of a circle is [latex]4.5[/latex] feet. Find its area. Use [latex]\pi = 3.1416[/latex].

[latex]15.90[/latex] square feet

Problem 3: Find the area of the circle below with a given radius. Use [latex]\pi = 3.14[/latex]

[latex]907.46[/latex] square centimeters

Problem 4: Find the area of the circle below with a given diameter. Use the value of [latex]\pi[/latex] on your calculator.

Make sure that you use the internal value of [latex]\pi[/latex] on your calculator.

[latex]60.00[/latex] square inches

Problem 5: The circumference of a circle is [latex]22.2[/latex] feet. What is its area? Use [latex]\pi = 3.14[/latex]

[latex]39.24[/latex] square feet

Problem 6: Determine the area of a dinner plate with a circumference of [latex]37.68[/latex] inches. Use [latex]\pi = 3.14[/latex].

[latex]113.04[/latex] square inches

Problem 7: The radius of a circle is [latex]5[/latex] inches. Find the area of the circle expressed in square centimeters [latex]c{m^2}[/latex]. Use 1 in = 2.54 cm. Use [latex]\pi = 3.14[/latex].

Convert [latex]5[/latex] inches to centimeters.

The area is

[latex]506.45[/latex] square centimeters

Problem 8: The diameter of a circle is [latex]12.4[/latex] miles. Calculate the area of the circle in terms of square kilometers [latex]k{m^2}[/latex]. Use 1 mi = 1.609 km. Use [latex]\pi = 3.1416[/latex].

Convert [latex]12.4[/latex] miles to kilometers.

[latex]312.64[/latex] square kilometers

Problem 9: What is the radius of a circle with an area of [latex]73.12[/latex] square miles [latex]m{i^2}[/latex]? Use [latex]\large{\pi = {{22} \over 7}}[/latex].

[latex]4.82[/latex] miles

Problem 10: Determine the diameter of the circle having an area of [latex]100[/latex] square yards [latex]y{d^2}[/latex]. Use [latex]\large{\pi = {{22} \over 7}}[/latex].

[latex]11.28[/latex] miles

Problem 11: Find the area of the semicircle below with a diameter of [latex]8[/latex] centimeters. Use [latex]\pi = 3.1416[/latex].

The area of a semicircle is half of the area of a circle.

[latex]27.33[/latex] square centimeters

Problem 12: Both circles share the same center. Find the exact area of the shaded region.

Area of the shaded region = area of the larger circle minus area of the smaller (inner) circle

[latex]21\pi [/latex] square inches

You may also be interested in these related math lessons or tutorials:

Area of a Circle

o          Circle

o          Radius

o          Diameter

o          Arc

o          Subtend

o          Sector

o          Chord

o          Secant

o          Tangent

o          Circumference

o          Know how to fundamentally define a circle

o          Identify various parts of circles

o          Calculate the area and circumference of a circle or portion thereof

Introduction to Circles

A circle is simply a figure defined by all the points that are equidistant from a given center point. Thus, we can define a circle (without a definite location) by specifying the distance from the center; alternatively, we can define a circle (with a definite location) by specifying the distance from and the location of the center. The distance from the center of the points that lie on the circle is called the radius (plural radii ) of the circle. A circle with a radius r and center point C is shown below.

Note that regardless of which point on the circle is chosen, it is always a distance r from the center point, C. Each of the line segments drawn from C to the circle is called a radius (in other words, no one segment is defined as the radius, since all such segments are equal in length). The circle itself does not show any angles or sides that we can use to determine how many degrees are in the figure (as we did with polygons), but we can see that any two radii do form an angle α , as shown below.

Using our measurement of degrees, this angle α can (unambiguously) be any value between 0° and 360°. We can also define angles with either positive or negative numbers, depending on the direction of measurement from a particular radius--a positive angle is traditionally measured in the counterclockwise direction, and a negative angle is traditionally measured in the clockwise direction, as shown below.

We can also identify other parts of a circle. Given any point P 1 on the circle, the most distant point P 2 that is also on the circle is on the exact opposite side, and these points are the length of two radii apart. The line segment that connects them passes through the center and is called the diameter of the circle.

Any two radii in the circle (which form an angle α ), such as those shown below, form an arc and a sector. The arc is the portion of the circle opposite to (or subtended by) the angle α and between the endpoints of the radii. The sector is the area enclosed by the arc and the radii. The arc A in the diagram below is shown as a bold line, and the sector S is shown as a shaded area.

Any line segment connecting two points on the circle is called a chord . Note that a diameter is a chord (the longest possible chord of a circle!). The chord D is shown below as a bold line.

Other special figures related to circles are secants and tangents. A secant is simply a line that intersects two points of the circle (a chord is a segment of a secant). A tangent is a line that intersects the circle at exactly one point. Secant E and tangent T are shown in the diagram below as bold lines.

Practice Problem : Identify each part (labeled A through E) of the circle below.

Solution : Each portion of the circle or other line can be identified by its relationship to the entire circle. Each of these figures is discussed and defined above. The shaded area A is a sector. The line (or line segment) B is a tangent (note that it intersects the circle at only one point). C is a diameter, D is a chord, and E is a radius.

Basic Characteristics of Circles

Now that we have identified some of the components of circles, we can now begin to derive some of their characteristics using the tools we have developed thus far. Some of the properties of circles require trigonometry to derive (such as the length of a chord subtended by an angle α ), but others we can derive or simply state basic formulas that we can use to solve problems. Let's start with the circumference and area of a circle. The circumference of a circle is simply the length of the boundary (that is, the perimeter) of the circle. We will simply state the formula for the circumference C of a circle in terms of the radius ( r ) or the diameter ( d ):

C = 2 πr = πd

Note that with the formula for the circumference, we introduce the number π . Because π is (most mathematicians think) an irrational number, we cannot write it out exactly in its decimal form, nor can we write it exactly as a fraction. We can, however, write a decimal estimation of π that is sufficient for our purposes.

π ≈ 3.1416

Many calculators have a built-in key for the number π (although the calculator still just uses an approximation of π ). For many calculations, an approximate π value of 3.14 yields sufficient accuracy. As it turns out, π also appears in the formula for the area of a circle. Let's write the area A in terms of the radius r .

A = πr 2

Note again that we have not really derived these formulas; we are simply stating them as fundamental facts upon which we will base the rest of our investigation of the characteristics of circles.

Practice Problem : Find the area and circumference of a circle with a diameter of 4 inches.

Solution : One of the first rules of solving these types of problems involving circles is to carefully note whether we are dealing with the radius or the diameter. In this problem, the circle is described using the diameter, which is 4 inches. The radius is thus 2 inches. Let's now calculate the area A and circumference C using the formulas given above.

A = πr 2 = π (2 in) 2 ≈ (3.14)(4 in 2 ) =12.56 in 2

C = πd = π (4 in) ≈ (3.14) (4 in) = 12.56 in

Coincidentally, the area and circumference have the same numerical values (but not the same units!). This is not generally the case, of course.

Let's take a look at the characteristics of other parts of a circle. For instance, now that we know how to calculate the circumference of a circle, we can also calculate the length of an arc (which is simply a portion of the circumference). An angle α defined by two radii subtends an arc. Let's take a look at several examples, from which we can identify a pattern. The arc K in each case is shown as a bold curve. The circumference of the circle is C .

The expressions under each example can be derived by inspection. We know that if the angle α is 90° (one-quarter of the full 360° of the circle), then the subtended arc is one-quarter of the circumference. The same reasoning applies to determining the arc length K for the other two cases as well. Thus, we can see that the arc length is related to the circumference as the angle α is related to 360°. But this is simply a ratio that we can write as follows.

Thus, the area of sector S is related to the area A according to the ratio of α to 360°. We can once again derive a formula.

Practice Problem : A central angle γ in a circle of radius 10 units forms a sector with an area of 2.62 square units. Find the measure of γ .

Solution : Let's start by drawing a diagram of the problem. This diagram need not be to scale-we can simply use it to more easily identify the parts of the circle discussed in the problem. We'll call the sector S , and we know S = 2.62 units 2 .

We can use the formula presented earlier to relate the area S of the sector to the area A of the circle.

But we also know how to calculate A using the given radius of 10 units.

A = π r 2 = π(10 units) 2 ≈ 3.14(100 units 2 ) ≈ 314 units 2

Now let's find γ .

Thus, the angle has a measure of 3°.

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Choose Your Test

Sat / act prep online guides and tips, circles on sat math: formulas, review, and practice.

Though triangles are far and away the most common geometric shape on the SAT, make sure not to underestimate the importance of circles. You will generally come across 2-3 questions on circles on any given SAT, so it’s definitely in your best interest to understand the ins and outs of how they work. And this guide is here to show you the way.

This will be your complete guide to SAT circles , including areas, circumferences, degrees, arcs, and points on a circle. We’ll take you through what these terms mean, how to manipulate and solve for various aspects of a circle, and how to tackle the most difficult SAT circle questions you may see on test day.

What Are Circles?

A circle is a two dimensional shape that is formed from the infinite number of points equidistant (the same distance) from a single point. This single point becomes the center of the circle.  

This means that any and all straight lines drawn from the circle’s center will exactly hit the edge of the circle, so long as all the lines are of equal length.

Though you can measure a circle in both degrees and radians, you will only ever have to use degrees on the SAT. Because of this, we will only be talking about degree measures in this guide.

A full circle has 360 degrees. A semicircle (half a circle) has $360/2 = 180$ degrees. This is why a straight line always measures 180 degrees.

To find a piece of a circle, you must find it in relation to 360 degrees. So a fifth of a circle is $360(1/5) = 72$ degrees, and an eighth of a circle is $360(1/8) = 45$ degrees, etc.

Circumference

The circumference is the edge of the circle. It is made from the infinite points equidistant from the center. 

In formulas, the circumference is represented as $c$. 

A diameter is any straight line drawn through the center of the circle that connects two opposite points on the circumference.

In formulas, the diameter is represented as $d$. 

The radius of a circle is a straight line drawn from the center of the circle to any point on the circumference. It is always half the diameter.

In formulas, the radius is represented as $r$. 

Circles are described as “tangent” with one another when they touch at exactly one point on each circumference.

A group of circles, all tangent to one another.  

If you’ve taken a geometry class, then you are also probably familiar with π (pi). π is the mathematical symbol that represents the ratio of any circle’s circumference to its diameter. It is usually expressed as 3.14(159), but its digits go on infinitely. (For more information on ratios, check out our guide to SAT ratios .)

Let's say we have a circle with a particular diameter (any diameter). Now let's multiply this same circle a few times and line them all up in a row. This gives us our same diameter 4 times in a line. 

Now, let us assign a starting point somewhere on the circumference of the circle and then "unpeel" the circumference from our circle.

Once you remove the circumference and lay it flat, you can see that the circumference is a little more than 3 full lengths of the circle's width/diameter (specifically, 3.14159 times).

The circumference of the circle will always be 3.14159 (π) times the diameter. So, if a circle’s diameter is 1, then its circumference is π. And if its diameter is 2, then its circumference is 2π, etc.

We can measure all the distance ever traveled (with wheels) in increments of pi. 

Circle Formulas

You will always be given a box of formulas on each SAT math section. This means it is not crucial for you to memorize circle formulas, but we still recommend that you do so if possible. Why? To help both your time management and problem solving ability.

The box of formulas you'll be given on every SAT math section. 

In terms of time management, memorizing your formulas will save you time from pulling up the formula sheet over and over again, which takes your attention away from the question you’re working on. And, on a timed standardized test like the SAT, every second counts.

It is also in your best interest to memorize your formulas simply for ease, practice, and familiarity. The more comfortable you get in knowing how circles work, the more quickly and easily you’ll be able to solve your problems.

So let’s look at your formulas.

$$c = 2πr$$

There are technically two formulas to find the circumference of a circle, but they mean exactly the same thing. (Why? Because any diameter will always be equal to twice the circle’s radius).

Because π is the relationship between a circle’s diameter and its circumference, you can always find a circle’s circumference as long as you know its diameter (or its radius) with these formulas. 

Here, we have two half circles and the sum of two radii, $RS = 12$.

We can either assign different values for the radius of circle R and the radius of circle S such that their sum is 12, or we can just mentally mash the two circles together and imagine that RS is actually the diameter of one circle. Let’s look at both methods.

Since we know that $RS = 12$, let us say that circle R has a radius of 4 and circle S has a radius of 8. (Why those numbers? Because all that matters is that the radii add up to equal 12. We could have picked 6 and 6, 10 and 2, 3 and 9, etc., so long as their sum was 12.)

So the circumference of circle R would be:

But, since we only have half a circle, we must divide that number in half.

${8π}/2 = 4π$

Now, we can do the same for circle S. But we can also see that it is a semi-circle. So instead of taking our circumference of $2πr$ for the whole circumference, let us just take the circumference of half ($πr$) and so save ourselves the trouble of all the steps we used for circle R.

${1/2}c = πr$

${1/2}c = 8π$

So now let us add our circumferences.

$4π + 8π = 12π$

So our final answer is C , $12π$

On the other hand, we could simply imagine that line RS is the diameter of a complete circle. (Why are we allowed to do this? Because we have the sum of two radii and two half circles, so combined, they would become one circle.)

If RS is a diameter of a circle whose complete circumference we must find, let us use our circumference formula.

$c = πd$ or $c = 2πr$

Again, our answer is C , $12π$.

$$a = πr^2$$

You can also use π to find the area of a circle as well, since a circle’s area is closely related to its circumference. (Why? A circle is made of infinite points, and so it is essentially made up of infinite triangular wedges--basically a pie with an infinite number of slices. The height of each of these wedges would be the circle’s radius and the cumulative bases would be the circle’s circumference.)

A circle splitting into a series of triangles.

So you would be able to find a circle’s area using the formula:

$$c \arc = πd({\arc \degree}/360°)$$

$$a \arc \sector = πr^2({\arc \degree}/360°)$$

In order to find the circumference of a circle’s arc (or the area of a wedge made from a particular arc), you must multiply your standard circle formulas by the fraction of the circle that the arc spans.

To determine the fraction of the circle that the arc spans, you must have the degree measure of the arc and find its measure out of the circle’s full 360 degrees. So if you want to find the circumference of an arc that is 90°, it would be $1/4$ the total area of the circle. Why? Because $360/90 = 4$ (in other words, $90/360 = 1/4$).  

This question gives us a lot of information, so let’s go through it piece by piece.

First of all, we are trying to find the length of an arc circumference, which means that we need two pieces of information--the arc degree measure and the radius (or the diameter).

Well, we have the degree measure, so we’re halfway there, but now we need the radius (or diameter) of the smaller circle. We are told that it is half the radius of the larger circle, so we must find the radius of the larger circle first.

All that we are told about the larger circle is that it has a circumference of 36. Luckily, we can find its radius from its circumference.

[Note: though it is unusual, this problem gives us our radius in pi units, rather than giving our circumference(s) in pi units. As we said, this is perfectly acceptable, though uncommon.]

If the circumference of the larger circle is 36, then its diameter equals $36/π$, which means that its radius equals $18/π$.

Because we know that the smaller circle has a radius that is half the length of the radius of the larger circle, we know that the radius of the smaller circle is:

$({18/π})/2 = 9/π$

So the radius of our smaller circle is $9/π$. This means we can finally find the arc measure of the smaller circle’s circumference, by using the radius of the circle and the interior degree measure.

$c_\arc = 2πr({\arc \degree}/360)$

$c_\arc = 2π({9/π})(80/360)$

$c_\arc = 4$

So our final answer is D , 4.  

The relationship between circles and pi is constant and unbreakable. 

Typical Circle Questions on the SAT

Circle problems on the SAT will almost always involve a diagram. With very rare exceptions, you will be given a picture from which to work. But we will discuss both diagram and word problems here on the chance that you will get multiple types of circle problems on your test.

Diagram Problem

A diagram problem will give you a diagram from which to work. You must use the visual you are provided and either find a missing piece or find equivalent measurements or differences.

Helpful hint: often (though not always), the trick to solving a circle problem is in finding and understanding the radius. All lines drawn from the center of the circle to the circumference are radii, and are therefore equal. This will often play a vital part to solving the whole problem.

Here is a perfect example of when the radius makes all the difference in a problem.

We are told that lines AB and AO are equal. Based on our knowledge of circles, we also know that AO and BO are equal. Why? Because they are both radii, and the radii of a circle are always equal.

This means that AB = AO = BO, which means that the triangle is equilateral.

Equilateral triangles have all equal sides and all equal angles, so the measure of all its interior angles are 60°. (For more on equilateral triangles, check out our guide to SAT triangles )

So angle measure ABO = 60 degrees.

Our final answer is D.

Word Problem

Word problem questions about circles will describe a scene or situation that revolves around circles in some way.

Generally, the reason why you will not be given a diagram on a circle question is because you are tasked with visualizing different types of circle types or scenarios. On rare occasions, you may get a word problem on circles because the question describes an inequality, which is difficult to show in a diagram.

When given a word problem question, it is a good idea to do your own quick sketch of the scene. This will help you keep all the details in order and/or see if you can make multiple types of shapes and scenarios, as with this question:

Here, we are being asked to visualize several potential different shapes and outcomes of this circle, which is why this problem is presented to us as a word problem. Because there are many different ways to draw out this scenario, let us look to the answer choices and either eliminate them or accept them as we go along.

Option I considers the possibility that M could be the center of the circle if lines XM and YM are equal and X and Y both lie somewhere on the circumference of the circle. We know this must be true because M being the center point of the circle would make lines XM and YM radii of the circle, which would mean that they were equal.

So option I is true and we can therefore eliminate answer choices B and D.

Now let’s look at option II.

Option II presents us with the possibility that point M lies somewhere on the arc of XY. Well, if point M rested exactly halfway between X and Y, then straight lines drawn from X to M and Y to M would certainly be equal.

So option II is also correct.

Finally, let’s look at option III.

Option III presents us with the possibility that M lies somewhere on the outside of the circle. So long as M lies at a distance halfway between X and Y, this scenario would still work.

So option III is also correct.

This means that all of our options (I, II, and III) are possible.

Our final answer is E.

Now let's talk circle tips and tricks.

How to Solve a Circle Problem

Now that you know your formulas, let’s walk through the SAT math tips and strategies for solving any circle problem that comes your way.

#1: Remember your formulas and/or know where to look for them

As we mentioned earlier, it is always best to remember your formulas when you can. But if you don’t feel comfortable memorizing formulas or you fear you will mix them up, don’t hesitate to look to your formula box--that is exactly why it is there.

Just be sure to look over the formula box before test day so that you know exactly what is on it, where to find it, and how you can use that information. (For more on the formulas you are given on the test, check out our guide to SAT math formulas .)

#2: Draw, draw, draw

If you’re not given a diagram, draw one yourself! It doesn’t take long to make your own picture and doing so can save you a lot of grief and struggle as you go through your test. It can be all too easy to make an assumption or mix up your numbers when you try to perform math in your head, so don’t be afraid to take a moment to draw your own pictures.

In Bluebook, you won’t be able to mark up diagrams in the actual exam. But the digital SAT features an online notepad that you can use to reproduce and mark up diagrams that are provided on exam questions . Alternatively, you can bring a pen or pencil, and the proctor will provide paper if you ask for it. The reason not everything is marked in your diagrams is so that the question won’t be too easy, so always write in your information yourself.

#3: Analyze what’s really being asked of you

All the formulas in the world won’t help you if you think you’re supposed to find the area, but you’re really being asked to find the circumference. Always remember that standardized tests are trying to get you to solve questions in ways in which you’re likely unfamiliar, so read carefully and pay close attention to the question you’re actually being asked.

#4: Use your formulas

Once you’ve verified what you’re supposed to find, most circle questions are fairly straightforward. Plug your givens into your formulas, isolate your missing information, and solve. Voila!

Test Your Knowledge

Now let's put your newfound circle knowledge to the test on some real SAT math problems. 

Answers: C, D, C

Answer Explanations:

1) This question involves a dash of creativity and is a perfect example of a time when you can and should draw on your given diagrams (had you been presented this on paper, that is).

We know that the inscribed figure is a square, which means that all of its sides are equal (for more on squares, check out our guide to SAT polygons ). Therefore, if you draw a line connecting points R and T, you will have a perfect semi-circle, or 180°.

Now, the arc we are looking for spans exactly half of that semi-circle.

This means that the arc degree measure of ST is:

$180/2 = 90$ degrees.

So our final answer C.

2) Now, before we even begin, read the question carefully. The question wants us to find the perimeter of the shaded region. If you were going too quickly through the test, you may have been tempted to find the area of the shaded region instead, which would have gotten you a completely different answer.

Because we are trying to find the perimeter of circular figures, we must use our formula for circumferences.

Let us start with the two circles in the middle.

We know that each circle has a radius of 3 and that our shaded perimeter spans exactly half of each circle. So the circumference for each small circle is:

And there are two small circles, so we must double this number:

$3π * 2 = 6π$

So the interior perimeter is $6π$.

Now, let’s find the outer perimeter, which is the circumference for half the larger circle.

If the radius of each of the small circles is 3, then that means the diameter of each small circle is:

$3 * 2 = 6$

And the diameter of each small circle is the same as the radius of the larger circle. This means that the full circumference of the larger circle is:

But we know that our perimeter only spans half the outer circumference, so we must divide this number in half.

${12π}/2 = 6π$

Our outer perimeter equals $6π$ and our inner perimeter equals $6π$. To get the full perimeter, we must add them together.

$6π + 6π = 12π$

Our final answer is D , $12π$.

3) Here, we are beginning with the understanding that the circle has an area of $25π$. We are tasked with finding the perimeter of one of the wedges, which requires us to know the radius length of the circle. This means we must work backwards from the circle’s area in order to find its radius.

Well the formula for the area of a circle is:

Our area equals 25, so:

Our radius measurement equals 5.

Now, we must find the arc measurement of each wedge. To do so, let us find the full circumference measurement and divide by the number of wedges (in this case, 8).

The full circumference is $10π$ which, divided by 8, is:

${10π}/8 = {5/4}π$

Now, let us add that arc measurement to twice the radius value of the circle in order to get the full perimeter of one of the wedges.

$5 + 5 + {5/4}π$

$10 + {5/4}π$

So our final answer is C.

The Take-Aways

Almost always, the most useful part of any circle will be the radius. Once you’ve gotten used to thinking that all radii are equal, then you will often be able to breeze past even the trickiest of SAT circle problems.

If you understand how radii work, and know your way around both a circle’s area and its circumference, then you will be well prepared for most any circle problem the SAT can dream up. Know that the SAT will present you with problems in strange ways, so remember your tricks and strategies for circle problems. Be careful with your work, keep a clear head, and you’ll do just fine.

What’s Next?

You've triumphed over circles (huzzah!). So now what? Well we've got guides aplenty on any SAT math topic you want to brush up on. Feel iffy on your lines and angles ? How about probability ? Integers ? Check out our SAT math tab on the blog for any SAT math topic questions you might have.

Don't know where to start? First, make sure you understand how the test is scored and what makes a "good" score or a "bad" score , so that you can figure out how you currently stack up.

Want to get a 600 on the SAT math? How about a perfect 800? Check out our articles on how to bring your scores up to a 600 and even how to get a perfect score on the SAT math, written by a perfect SAT-scorer.

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Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time.

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Godowski, M.Ed.

• Dec 18, 2021

RESTORATIVE PROBLEM SOLVING & BRAINSTORMING CIRCLE

Use this Circle Facilitator Guide to discuss needs, problem solving, and feedback within your community, team, school, organization, or group and practice a method for Restorative problem solving and brainstorming using a fishbowl-style circle.

Welcome to the Space of the Circle.

MINDFULNESS MOMENT Pause, breathe, and listen to the sound

OPENING Land Acknowledgement Visit https://native-land.ca/ for more info.

INTRODUCE "ROUNDS" A "round" is a pass of the talking piece around the Circle. The keeper poses a question and, as a participant, may answer first. The keeper then passes the talking piece to the person to their left or right (or names the next person over Zoom). On the first round, participants are invited to say their name and pronouns as well as respond to the question. Remember, it is always okay for a participant to pass.

Remind the group of Circle values and guidelines. These can be ones established in previous circles or be created new each time with new participants. Some helpful sharing guidelines may include:

Share what you feel comfortable sharing.

Respect the talking piece

You can pass at any time.

Be mindful of time.

Respect the privacy of those sharing.

Trust intent/name impact

Use “I” statements

Speak from the heart/listen from the heart

Be here now

Prioritize connection before content

Be both teachers and learners

ROUND Why is it important that you are here right now?

ROUND Share a current need or problem within your community (organization, team, school, personal, etc.).

ROUND Describe a way your community currently works to (re)solve problems or generate new ideas.

ROUND How do you seek out feedback?

ROUND How do you respond to feedback?

LEARNING ACTIVITY Review concepts of circle practices and introduce fishbowl circles

RESTORATIVE PROBLEM SOLVING CIRCLE

Presentation of a problem or issue - 3 interrupted minutes to describe the problem and let the group know the need

Group has 10 minutes to give feedback. Feedback should be in the form of a declarative statement, not a question.

The group may ask clarifying questions, but this should not be the focus

The volunteer does not interrupt, just listens and take notes

Volunteer has 2 minutes to reflect on what they heard and let the group know one or two things they are going to try

CHECK-OUT ROUND Share one word or a short phrase for how you are feeling right now?

The truth is that you cannot build community without conflict. The issue is not to be without conflict, but to be able to resolve conflict, and the commitment to community is what gives us the inspiration to come up with ways to resolve conflict.

- bell hooks

“The Beloved Community”

Thank everyone for participating in the Circle!

Problem-Solving Circle adapted from Professional learning group talking and facilitation points. (2015). International Institute for Restorative Practices.

#needs #problems #ideas #restorative #circle #facilitatorguide #problemsolving #brainstorming #fishbowl #values #community

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RESTORATIVE QUESTIONS

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Tangent to Circles - Problem Solving

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• Satyajit Ghosh

One tangent line, and only one, can be drawn to any point on the circumference of a circle, and this tangent is perpendicular to the radius through the point of contact. In the figure below, line \(BC\) is tangent to the circle at point \(A\).

If two tangents are drawn to a circle from an external point,

• the lengths of the tangents from the external point to the points of contact are equal;
• they subtend equal angles at the center of the circle.

In the figure below, \(AB=CB, \angle AOB= \angle COB\) and \(\angle ABO=\angle CBO\).

In the figure below, \(BA\) and \(BC\) are tangents to the circle at points \(A\) and \(C\), respectively. If \(\angle AEC=110^\circ\) and \(\angle DCE=80^\circ\), find \(\angle ABC\). SOLUTION Consider the diagram to the right. Opposite angles of a cyclic quadrilateral sum up to \(180^\circ\). So \[\angle ADC=180^\circ-110^\circ=70^\circ.\] From the inscribed angle theorem , we can see that \[\angle AOC=2 \angle ADC=2(70^\circ)=140^\circ.\] Since the sum of interior angle of a quadrilateral is \(360^\circ\), we have \[\begin{align} \angle ABC &=360^\circ-\angle BCO-\angle BAO-\angle AOC\\ &=360^\circ-90^\circ-90^\circ-140^\circ\\ &=40^\circ.\ _\square \end{align}\]

On the same side of a straight line three circles are drawn as follows:

• A circle with a radius of 4 cm is tangent to the line.
• The other two circles are identical, and each is tangent to the line and to the other two circles.

What are the possible radii of these two identical circles?

A.\(\ \, \) 24 cm only B.\(\ \, \) 20 cm only C.\(\ \, \)16 cm only D.\(\ \, \) 24 and 20 cm only E.\(\ \, \) 20 and 16 cm only F.\(\ \, \) 16 and 24 cm only G.\(\ \, \) 16, 20, 14 cm only H.\(\ \ \) No solution

\(\) Note: A diagram has not been provided because drawing the correct diagram is part of the problem solving.

In the figure to the right, the yellow circle is tangent to \( \triangle ABC \) at points \( D, E, F. \)

\( 2 \sqrt{14}, 4 \sqrt{3}, \) and \( \sqrt{42} \) are the lengths of the common tangents to the semicircles of diameters \((BD, CD), (CE, AE),\) and \((AF, BF),\) respectively.

What is the radius of the yellow circle?

In the diagram below, line segments \(STM\) and \(MQ\) are tangents to the circle at \(T\) and \(Q,\) respectively. \(C\) is the center of the circle and does not lie on \(QS.\) If \(SRQ\) is a straight line, \(SR=TR,\) and \(\angle TSR=25^\circ,\) what is \(\angle QMT\) in degrees?

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• Math Article
• Problems On Circles

Problems On Area And Circumference Of A Circle

Before looking at problems on a circle based on perimeter and area, we need to understand the meaning of a circle and both the properties of the circle. In geometry, a circle is a closed two-dimensional figure in which the set of all the points in the plane is equidistant from a given point called the centre. Let’s have a look the definitions of perimeter and area given below:

Perimeter Definition

Perimeter is associated with any closed figure like triangle , quadrilateral, polygons or circles. It is the measure of distance covered while going around the closed figure on its boundary.

For example, the perimeter of a square of side 2 cm = 8 cm, as we know that the square comprises 4 sides having equal lengths, thus the total distance covered will be 4×2, which will be the total length (i.e. perimeter).

Area Definition

Area means the actual space enclosed by a closed figure (or within the perimeter), and it means all the points within the closed figure and not the boundary.

Perimeter and Area of Circle Formula

Now, coming to the perimeter of a circle, as explained above, is the measure of distance going around the circle’s boundary. This distance is difficult to calculate precisely, and it is easy to visualize that the distance to go around is larger in a larger radius than a smaller radius. Hence, the perimeter is a function of the radius of a circle. In the case of the circle, we generally use the term “CIRCUMFERENCE” instead of the perimeter.

It is given by,

Circumference = 2πr

Here, r is a radius, and π is a constant defined as the ratio of circumference to the circle’s diameter.

The value of π is 22/7 or 3.1416

Also check: Circumference of a circle

The formula to find the area of a circle is given as:

Solved Problems on Circle

Let us understand the concepts related to circles along with the following questions-

To cover a distance of 10 km, a wheel rotates 5000 times. Find the radius of the wheel.

Number of rotations = 5000.

Total distance covered = 10 km

Let ‘r’ be the radius of the wheel.

Circumference of the wheel = Distance covered in 1 rotation = 2πr.

In 5000 rotations, the distance covered = 10 km = 1000000 cm.

\(\begin{array}{l}\text{Hence, in 1 rotation, the distance covered = }\frac{1000000}{5000}cm=200\: cm\end{array} \)

But this is equal to the circumference. Hence, 2πr = 200 cm

Taking the approximate value of π as 22/7, we get

r = 100 x 7/22

r = 31.82 cm approx.

The diameter of a semicircular shape is 14 cm. What will be the perimeter of this shape?

Diameter of semicircle = d = 14 cm

Radius = r = d/2 = 14/2 = 7 cm

Perimeter of semicircle = (Perimeter of circle/2) + d

= (2πr/2) + d

= (22/7) × 7 + 14

The difference between the circumference and the diameter of a circular bangle is 5 cm. Find the radius of the bangle. (Take π = 22/7)

Solution: Let the radius of the bangle be ’r’

According to the question:

Circumference – Diameter=5 cm

We know, Circumference of a circle = 2πr

Diameter of a circle = 2r

Therefore, 2πr – 2r =5 cm

2r(π-1) = 5 cm

\(\begin{array}{l}2r(\frac{22}{7}-1)=5cm\\ \\ 2r\times \frac{15}{7}=5\\ \\ r=\frac{5\times 7}{15\times 2}\\ \\ r=1.166cm\end{array} \)

The radius of bangle is 1.166 cm.

A girl wants to make a square-shaped figure from a circular wire of radius 49 cm. Determine the sides of a square.

Solution: Let the radius of the circle be ’r’.

Length of the wire=circumference of the circle= 2πr

\(\begin{array}{l}= 2\times \frac{22}{7}\times 49=2\times 22\times 7\\ \\ =308\: cm\end{array} \)

Let the side of the square be ‘s’.

Perimeter of the square = length of the wire = 4s

\(\begin{array}{l}s=\frac{308}{4}\\ \\ s=77\:cm\end{array} \)

Therefore, the sides of the square is 77 cm.

Find the area of a circular region whose radius is 21 m.

Radius of circular region = r = 21 m

Area of a circle = πr 2

= (22/7) × 21 × 21

= 22 × 3 × 21

= 1386 sq. m

Therefore, the area of the circular region is 1386 sq. m.

Video Lessons on Circles

Introduction to circles.

Parts of a Circle

Area of a Circle

All about Circles

Learn more about Surface area & volumes of the figures with BYJU’S-The learning app.

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• Integrations
• Learning Center

CIRCLES Method

What is the circles method.

The CIRCLES method is a problem-solving framework that helps product managers (PMs) make a thorough and thoughtful response to any design question.

The seven linear steps of the process form the CIRCLES acronym: C omprehend the situation; i dentify the customer; r eport the customer’s needs; c ut, through prioritization; l ist solutions; e valuate tradeoffs, and s ummarize your recommendation.

Key Concepts of the CIRCLES Method

The sequential structure of the CIRCLES method enables PMs to move through essential questions to understand what needs to design and why fully. Some consider the CIRCLES method to be a checklist for asking the right questions when forming an exhaustive and organized response to a design question.

According to Lewis C. Lin, author of Decode and Conquer and creator of the CIRCLES method, the first critical step — comprehending the situation — is a three-fold process that involves:

• Clarifying the goal (e.g., increase revenue, market share, or engagement).
• Understanding the constraints you have for the problem upfront (e.g., how much time do you have, how many engineer resources are available, etc.).
• Understanding the context of the situation that gives you foundational knowledge (i.e., don’t guess or make assumptions–instead ask questions that help you understand, like “What is it?” and “Who is it for?”).

Hear Lewis C. Lin walk through the CIRCLES method .

Here are the seven steps to the CIRCLES method:

Alicia Newman at Learn Worthy writes:

“The CIRCLES framework is put together so that you can use mental cues to structure your response to a product design question. Knowing the backbone of the framework ensures that once you get that product design question, you will know which elements to include in your answer no matter what the product is.”

Why Is the CIRCLES Method Important to Product Management?

The CIRCLES method is useful in product management because it:

• Keeps the focus on users by distilling who PMs are building the product or feature for. Communicates why they are building it.
• Helps PMs prioritize things like product features, execution, user feedback, and the product roadmap
• Enables PMs to ask the right questions during the critical first step (comprehend the situation) to gather ample information before rushing to a solution
• Encourages PMs to keep an open mind as they move through the sequential steps of the framework instead of jumping to conclusions or a solution

Related terms: Customer Empathy /  User Experience /  Product Design /  Product Discovery /  Design Thinking

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11.1 Distance and Midpoint Formulas; Circles

Learning objectives.

By the end of this section, you will be able to:

Use the Distance Formula

Use the Midpoint Formula

• Write the equation of a circle in standard form
• Graph a circle

Be Prepared 11.1

Before you get started, take this readiness quiz.

Find the length of the hypotenuse of a right triangle whose legs are 12 and 16 inches. If you missed this problem, review Example 2.34 .

Be Prepared 11.2

Factor: x 2 − 18 x + 81 . x 2 − 18 x + 81 . If you missed this problem, review Example 6.24 .

Be Prepared 11.3

Solve by completing the square: x 2 − 12 x − 12 = 0 . x 2 − 12 x − 12 = 0 . If you missed this problem, review Example 9.22 .

In this chapter we will be looking at the conic sections, usually called the conics, and their properties. The conics are curves that result from a plane intersecting a double cone—two cones placed point-to-point. Each half of a double cone is called a nappe.

There are four conics—the circle , parabola , ellipse , and hyperbola . The next figure shows how the plane intersecting the double cone results in each curve.

Each of the curves has many applications that affect your daily life, from your cell phone to acoustics and navigation systems. In this section we will look at the properties of a circle.

We have used the Pythagorean Theorem to find the lengths of the sides of a right triangle. Here we will use this theorem again to find distances on the rectangular coordinate system. By finding distance on the rectangular coordinate system, we can make a connection between the geometry of a conic and algebra—which opens up a world of opportunities for application.

Our first step is to develop a formula to find distances between points on the rectangular coordinate system. We will plot the points and create a right triangle much as we did when we found slope in Graphs and Functions . We then take it one step further and use the Pythagorean Theorem to find the length of the hypotenuse of the triangle—which is the distance between the points.

Example 11.1

Use the rectangular coordinate system to find the distance between the points ( 6 , 4 ) ( 6 , 4 ) and ( 2 , 1 ) . ( 2 , 1 ) .

Try It 11.1

Use the rectangular coordinate system to find the distance between the points ( 6 , 1 ) ( 6 , 1 ) and ( 2 , −2 ) . ( 2 , −2 ) .

Try It 11.2

Use the rectangular coordinate system to find the distance between the points ( 5 , 3 ) ( 5 , 3 ) and ( −3 , −3 ) . ( −3 , −3 ) .

The method we used in the last example leads us to the formula to find the distance between the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) . ( x 2 , y 2 ) .

When we found the length of the horizontal leg we subtracted 6 − 2 6 − 2 which is x 2 − x 1 . x 2 − x 1 .

When we found the length of the vertical leg we subtracted 4 − 1 4 − 1 which is y 2 − y 1 . y 2 − y 1 .

If the triangle had been in a different position, we may have subtracted x 1 − x 2 x 1 − x 2 or y 1 − y 2 . y 1 − y 2 . The expressions x 2 − x 1 x 2 − x 1 and x 1 − x 2 x 1 − x 2 vary only in the sign of the resulting number. To get the positive value-since distance is positive- we can use absolute value. So to generalize we will say | x 2 − x 1 | | x 2 − x 1 | and | y 2 − y 1 | . | y 2 − y 1 | .

In the Pythagorean Theorem, we substitute the general expressions | x 2 − x 1 | | x 2 − x 1 | and | y 2 − y 1 | | y 2 − y 1 | rather than the numbers.

This is the Distance Formula we use to find the distance d between the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) . ( x 2 , y 2 ) .

Distance Formula

The distance d between the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) ( x 2 , y 2 ) is

Example 11.2

Use the Distance Formula to find the distance between the points ( −5 , −3 ) ( −5 , −3 ) and ( 7 , 2 ) . ( 7 , 2 ) .

Try It 11.3

Use the Distance Formula to find the distance between the points ( −4 , −5 ) ( −4 , −5 ) and ( 5 , 7 ) . ( 5 , 7 ) .

Try It 11.4

Use the Distance Formula to find the distance between the points ( −2 , −5 ) ( −2 , −5 ) and ( −14 , −10 ) . ( −14 , −10 ) .

Example 11.3

Use the Distance Formula to find the distance between the points ( 10 , −4 ) ( 10 , −4 ) and ( −1 , 5 ) . ( −1 , 5 ) . Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Try It 11.5

Use the Distance Formula to find the distance between the points ( −4 , −5 ) ( −4 , −5 ) and ( 3 , 4 ) . ( 3 , 4 ) . Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Try It 11.6

Use the Distance Formula to find the distance between the points ( −2 , −5 ) ( −2 , −5 ) and ( −3 , −4 ) . ( −3 , −4 ) . Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

It is often useful to be able to find the midpoint of a segment. For example, if you have the endpoints of the diameter of a circle, you may want to find the center of the circle which is the midpoint of the diameter. To find the midpoint of a line segment, we find the average of the x -coordinates and the average of the y -coordinates of the endpoints.

Midpoint Formula

The midpoint of the line segment whose endpoints are the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) ( x 2 , y 2 ) is

To find the midpoint of a line segment, we find the average of the x -coordinates and the average of the y -coordinates of the endpoints.

Example 11.4

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are ( −5 , −4 ) ( −5 , −4 ) and ( 7 , 2 ) . ( 7 , 2 ) . Plot the endpoints and the midpoint on a rectangular coordinate system.

Try It 11.7

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are ( −3 , −5 ) ( −3 , −5 ) and ( 5 , 7 ) . ( 5 , 7 ) . Plot the endpoints and the midpoint on a rectangular coordinate system.

Try It 11.8

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are ( −2 , −5 ) ( −2 , −5 ) and ( 6 , −1 ) . ( 6 , −1 ) . Plot the endpoints and the midpoint on a rectangular coordinate system.

Both the Distance Formula and the Midpoint Formula depend on two points, ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) . ( x 2 , y 2 ) . It is easy to confuse which formula requires addition and which subtraction of the coordinates. If we remember where the formulas come from, it may be easier to remember the formulas.

Write the Equation of a Circle in Standard Form

As we mentioned, our goal is to connect the geometry of a conic with algebra. By using the coordinate plane, we are able to do this easily.

We define a circle as all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, ( h , k ) , ( h , k ) , and the fixed distance is called the radius , r , of the circle.

A circle is all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center , ( h , k ) , ( h , k ) , and the fixed distance is called the radius , r , of the circle.

This is the standard form of the equation of a circle with center, ( h , k ) , ( h , k ) , and radius, r .

Standard Form of the Equation a Circle

The standard form of the equation of a circle with center, ( h , k ) , ( h , k ) , and radius, r , is

Example 11.5

Write the standard form of the equation of the circle with radius 3 and center ( 0 , 0 ) . ( 0 , 0 ) .

Try It 11.9

Write the standard form of the equation of the circle with a radius of 6 and center ( 0 , 0 ) . ( 0 , 0 ) .

Try It 11.10

Write the standard form of the equation of the circle with a radius of 8 and center ( 0 , 0 ) . ( 0 , 0 ) .

In the last example, the center was ( 0 , 0 ) . ( 0 , 0 ) . Notice what happened to the equation. Whenever the center is ( 0 , 0 ) , ( 0 , 0 ) , the standard form becomes x 2 + y 2 = r 2 . x 2 + y 2 = r 2 .

Example 11.6

Write the standard form of the equation of the circle with radius 2 and center ( −1 , 3 ) . ( −1 , 3 ) .

Try It 11.11

Write the standard form of the equation of the circle with a radius of 7 and center ( 2 , −4 ) . ( 2 , −4 ) .

Try It 11.12

Write the standard form of the equation of the circle with a radius of 9 and center ( −3 , −5 ) . ( −3 , −5 ) .

In the next example, the radius is not given. To calculate the radius, we use the Distance Formula with the two given points.

Example 11.7

Write the standard form of the equation of the circle with center ( 2 , 4 ) ( 2 , 4 ) that also contains the point ( −2 , 1 ) . ( −2 , 1 ) .

The radius is the distance from the center to any point on the circle so we can use the distance formula to calculate it. We will use the center ( 2 , 4 ) ( 2 , 4 ) and point ( −2 , 1 ) ( −2 , 1 )

Now that we know the radius, r = 5 , r = 5 , and the center, ( 2 , 4 ) , ( 2 , 4 ) , we can use the standard form of the equation of a circle to find the equation.

Try It 11.13

Write the standard form of the equation of the circle with center ( 2 , 1 ) ( 2 , 1 ) that also contains the point ( −2 , −2 ) . ( −2 , −2 ) .

Try It 11.14

Write the standard form of the equation of the circle with center ( 7 , 1 ) ( 7 , 1 ) that also contains the point ( −1 , −5 ) . ( −1 , −5 ) .

Graph a Circle

Any equation of the form ( x − h ) 2 + ( y − k ) 2 = r 2 ( x − h ) 2 + ( y − k ) 2 = r 2 is the standard form of the equation of a circle with center, ( h , k ) , ( h , k ) , and radius, r. We can then graph the circle on a rectangular coordinate system.

Note that the standard form calls for subtraction from x and y . In the next example, the equation has x + 2 , x + 2 , so we need to rewrite the addition as subtraction of a negative.

Example 11.8

Find the center and radius, then graph the circle: ( x + 2 ) 2 + ( y − 1 ) 2 = 9 . ( x + 2 ) 2 + ( y − 1 ) 2 = 9 .

Try It 11.15

ⓐ Find the center and radius, then ⓑ graph the circle: ( x − 3 ) 2 + ( y + 4 ) 2 = 4 . ( x − 3 ) 2 + ( y + 4 ) 2 = 4 .

Try It 11.16

ⓐ Find the center and radius, then ⓑ graph the circle: ( x − 3 ) 2 + ( y − 1 ) 2 = 16 . ( x − 3 ) 2 + ( y − 1 ) 2 = 16 .

To find the center and radius, we must write the equation in standard form. In the next example, we must first get the coefficient of x 2 , y 2 x 2 , y 2 to be one.

Example 11.9

Find the center and radius and then graph the circle, 4 x 2 + 4 y 2 = 64 . 4 x 2 + 4 y 2 = 64 .

Try It 11.17

ⓐ Find the center and radius, then ⓑ graph the circle: 3 x 2 + 3 y 2 = 27 3 x 2 + 3 y 2 = 27

Try It 11.18

ⓐ Find the center and radius, then ⓑ graph the circle: 5 x 2 + 5 y 2 = 125 5 x 2 + 5 y 2 = 125

If we expand the equation from Example 11.8 , ( x + 2 ) 2 + ( y − 1 ) 2 = 9 , ( x + 2 ) 2 + ( y − 1 ) 2 = 9 , the equation of the circle looks very different.

This form of the equation is called the general form of the equation of the circle .

General Form of the Equation of a Circle

The general form of the equation of a circle is

If we are given an equation in general form, we can change it to standard form by completing the squares in both x and y . Then we can graph the circle using its center and radius.

Example 11.10

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 4 x − 6 y + 4 = 0 . x 2 + y 2 − 4 x − 6 y + 4 = 0 .

We need to rewrite this general form into standard form in order to find the center and radius.

Try It 11.19

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 6 x − 8 y + 9 = 0 . x 2 + y 2 − 6 x − 8 y + 9 = 0 .

Try It 11.20

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 + 6 x − 2 y + 1 = 0 . x 2 + y 2 + 6 x − 2 y + 1 = 0 .

In the next example, there is a y -term and a y 2 y 2 -term. But notice that there is no x -term, only an x 2 x 2 -term. We have seen this before and know that it means h is 0. We will need to complete the square for the y terms, but not for the x terms.

Example 11.11

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 + 8 y = 0 . x 2 + y 2 + 8 y = 0 .

Try It 11.21

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 2 x − 3 = 0 . x 2 + y 2 − 2 x − 3 = 0 .

Try It 11.22

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 12 y + 11 = 0 . x 2 + y 2 − 12 y + 11 = 0 .

Access these online resources for additional instructions and practice with using the distance and midpoint formulas, and graphing circles.

• Distance-Midpoint Formulas and Circles
• Finding the Distance and Midpoint Between Two Points
• Completing the Square to Write Equation in Standard Form of a Circle

Section 11.1 Exercises

Practice makes perfect.

In the following exercises, find the distance between the points. Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

( 2 , 0 ) ( 2 , 0 ) and ( 5 , 4 ) ( 5 , 4 )

( −4 , −3 ) ( −4 , −3 ) and ( 2 , 5 ) ( 2 , 5 )

( −4 , −3 ) ( −4 , −3 ) and ( 8 , 2 ) ( 8 , 2 )

( −7 , −3 ) ( −7 , −3 ) and ( 8 , 5 ) ( 8 , 5 )

( −1 , 4 ) ( −1 , 4 ) and ( 2 , 0 ) ( 2 , 0 )

( −1 , 3 ) ( −1 , 3 ) and ( 5 , −5 ) ( 5 , −5 )

( 1 , −4 ) ( 1 , −4 ) and ( 6 , 8 ) ( 6 , 8 )

( −8 , −2 ) ( −8 , −2 ) and ( 7 , 6 ) ( 7 , 6 )

( −3 , −5 ) ( −3 , −5 ) and ( 0 , 1 ) ( 0 , 1 )

( −1 , −2 ) ( −1 , −2 ) and ( −3 , 4 ) ( −3 , 4 )

( 3 , −1 ) ( 3 , −1 ) and ( 1 , 7 ) ( 1 , 7 )

( −4 , −5 ) ( −4 , −5 ) and ( 7 , 4 ) ( 7 , 4 )

In the following exercises, ⓐ find the midpoint of the line segments whose endpoints are given and ⓑ plot the endpoints and the midpoint on a rectangular coordinate system.

( 0 , −5 ) ( 0 , −5 ) and ( 4 , −3 ) ( 4 , −3 )

( −2 , −6 ) ( −2 , −6 ) and ( 6 , −2 ) ( 6 , −2 )

( 3 , −1 ) ( 3 , −1 ) and ( 4 , −2 ) ( 4 , −2 )

( −3 , −3 ) ( −3 , −3 ) and ( 6 , −1 ) ( 6 , −1 )

In the following exercises, write the standard form of the equation of the circle with the given radius and center ( 0 , 0 ) . ( 0 , 0 ) .

Radius: 2 2

Radius: 5 5

In the following exercises, write the standard form of the equation of the circle with the given radius and center

Radius: 1, center: ( 3 , 5 ) ( 3 , 5 )

Radius: 10, center: ( −2 , 6 ) ( −2 , 6 )

Radius: 2.5 , 2.5 , center: ( 1.5 , −3.5 ) ( 1.5 , −3.5 )

Radius: 1.5 , 1.5 , center: ( −5.5 , −6.5 ) ( −5.5 , −6.5 )

For the following exercises, write the standard form of the equation of the circle with the given center with point on the circle.

Center ( 3 , −2 ) ( 3 , −2 ) with point ( 3 , 6 ) ( 3 , 6 )

Center ( 6 , −6 ) ( 6 , −6 ) with point ( 2 , −3 ) ( 2 , −3 )

Center ( 4 , 4 ) ( 4 , 4 ) with point ( 2 , 2 ) ( 2 , 2 )

Center ( −5 , 6 ) ( −5 , 6 ) with point ( −2 , 3 ) ( −2 , 3 )

In the following exercises, ⓐ find the center and radius, then ⓑ graph each circle.

( x + 5 ) 2 + ( y + 3 ) 2 = 1 ( x + 5 ) 2 + ( y + 3 ) 2 = 1

( x − 2 ) 2 + ( y − 3 ) 2 = 9 ( x − 2 ) 2 + ( y − 3 ) 2 = 9

( x − 4 ) 2 + ( y + 2 ) 2 = 16 ( x − 4 ) 2 + ( y + 2 ) 2 = 16

( x + 2 ) 2 + ( y − 5 ) 2 = 4 ( x + 2 ) 2 + ( y − 5 ) 2 = 4

x 2 + ( y + 2 ) 2 = 25 x 2 + ( y + 2 ) 2 = 25

( x − 1 ) 2 + y 2 = 36 ( x − 1 ) 2 + y 2 = 36

( x − 1.5 ) 2 + ( y + 2.5 ) 2 = 0.25 ( x − 1.5 ) 2 + ( y + 2.5 ) 2 = 0.25

( x − 1 ) 2 + ( y − 3 ) 2 = 9 4 ( x − 1 ) 2 + ( y − 3 ) 2 = 9 4

x 2 + y 2 = 64 x 2 + y 2 = 64

x 2 + y 2 = 49 x 2 + y 2 = 49

2 x 2 + 2 y 2 = 8 2 x 2 + 2 y 2 = 8

6 x 2 + 6 y 2 = 216 6 x 2 + 6 y 2 = 216

In the following exercises, ⓐ identify the center and radius and ⓑ graph.

x 2 + y 2 + 2 x + 6 y + 9 = 0 x 2 + y 2 + 2 x + 6 y + 9 = 0

x 2 + y 2 − 6 x − 8 y = 0 x 2 + y 2 − 6 x − 8 y = 0

x 2 + y 2 − 4 x + 10 y − 7 = 0 x 2 + y 2 − 4 x + 10 y − 7 = 0

x 2 + y 2 + 12 x − 14 y + 21 = 0 x 2 + y 2 + 12 x − 14 y + 21 = 0

x 2 + y 2 + 6 y + 5 = 0 x 2 + y 2 + 6 y + 5 = 0

x 2 + y 2 − 10 y = 0 x 2 + y 2 − 10 y = 0

x 2 + y 2 + 4 x = 0 x 2 + y 2 + 4 x = 0

x 2 + y 2 − 14 x + 13 = 0 x 2 + y 2 − 14 x + 13 = 0

Writing Exercises

Explain the relationship between the distance formula and the equation of a circle.

Is a circle a function? Explain why or why not.

In your own words, state the definition of a circle.

In your own words, explain the steps you would take to change the general form of the equation of a circle to the standard form.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

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Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction

• Authors: Lynn Marecek, Andrea Honeycutt Mathis
• Publisher/website: OpenStax
• Book title: Intermediate Algebra 2e
• Publication date: May 6, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/11-1-distance-and-midpoint-formulas-circles

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A book by David Hutchins

7 Basic Tools for Quality Circle Improvement and Management

Quality Circles are groups of individuals within an organization who come together to identify and solve problems, aiming for continuous improvement. These circles employ a set of effective techniques known as the Seven Basic Tools.

In this post, we will explore these tools and their significance in the Quality Circle improvement and management journey.

Tools for Quality Circle Improvement

Quality Circles are equipped with a fundamental set of problem-solving tools known as the Seven Basic Tools. These tools are the foundation on which Circles build their problem-solving capabilities, and as they mature, they may incorporate additional techniques to suit their evolving needs.

Problem-Solving Techniques

Effective problem-solving is at the core of Quality Circle activities. The process involves:

• Identifying problems.
• Gathering and analysing relevant data.
• Exploring potential causes.
• Proposing solutions.
• Evaluating the effectiveness of the solutions.
• Implementing them.
• Sustaining the improvements achieved.

Ensuring accuracy also requires a disciplined approach that relies on facts rather than opinions. Root Cause Analysis is a crucial aspect of this problem-solving discipline, allowing teams to pinpoint the underlying reasons for issues and avoid mere symptomatic treatment.

Overcoming Misconceptions

One common concern among managers unfamiliar with Quality Circles is the belief that unskilled or semi-skilled workers may not achieve impressive results. This misconception stems from underestimating the power of the simple yet effective techniques employed by Circles and overlooking the untapped potential of their members.

A dialogue often unfolds between managers and Circle members, where the managers ask why certain information wasn’t shared earlier, and the response is that they were never asked. This highlights the need to recognize all employees’ valuable insights and expertise.

Evolution of the Seven Basic Tools

The Seven Basic Tools were not initially designed for Quality Circles specifically. Instead, they gradually found their way into Circles during their formative years in Japan through trial and error. These tools proved particularly effective in small group activities, prompting their adoption as the essential problem-solving toolkit.

As Circles progress and mature, their members may seek to utilize these techniques in more sophisticated ways and incorporate additional skills relevant to their work, making the possibilities for growth endless.

Self-Managing Workgroups

As Quality Circles evolve, they have the potential to transform into self-managing workgroups. This entails an advanced level of autonomy and responsibility, where teams take charge of their decision-making processes and performance management.

Achieving self-managing workgroup status is a gradual process that requires dedication and alignment with organizational goals. While it may take time, the rewards in terms of efficiency, innovation, and employee satisfaction are well worth the effort. In the book “ Self Managing Workgroups” by David Hutchins , you will fully understand self managing workgroups as a key to participative management.

Integration with Six Sigma

Six Sigma is a well-known and sophisticated process improvement and quality management approach.

While some might assume that Six Sigma is beyond the scope of Quality Circles, it is crucial to recognize that there is no inherent limitation preventing Circles from embracing Six Sigma principles.

There are instances where the principles of Six Sigma can complement and enhance the efforts of Quality Circles, leading to even more impactful results.

The Seven Basic Tools in Detail

The foundation of Quality Circles lies in the application of the Seven Basic Tools. Let’s briefly explore each tool:

• Brainstorming: A creative technique for generating ideas and solutions through open discussion.
• Data Collection: Gathering relevant data to understand the scope and nature of the problem.
• Data Analysis: Analyzing the collected data to derive meaningful insights.
• Pareto Analysis: Prioritizing problems and focusing efforts on the vital few rather than the trivial many.
• Cause and Effect Analysis: Identifying the root causes of issues and understanding their relationships.
• Histograms: Utilizing graphical representations to visualize data distributions.
• Control Techniques: Implementing methods to monitor and sustain the improvements achieved.

To gain a deeper understanding of these tools, readers are encouraged to explore the book “ Self Managing Workgroups” by David Hutchins , which delves into the strategic aspects of implementing Quality circle in any industry.

In conclusion, quality circles leverage the power of the seven basic tools to drive continuous improvement and problem-solving. These fundamental techniques lay the groundwork for Circles to evolve into self-managing workgroups, transcending limitations and achieving remarkable outcomes. Integrating with other methodologies like Six Sigma, Quality Circles can further enhance their impact, contributing to the organization’s overall success.

Are the Seven Basic Tools exclusive to Quality Circles?

While these tools are closely associated with Quality Circles, they can be employed by any team or organization seeking effective problem-solving and continuous improvement.

How can managers foster a culture of problem-solving in their workplace?

Encouraging open communication, valuing employee input, and providing the necessary training and resources can foster a problem-solving culture.

What benefits do self-managing workgroups offer to organizations?

Self-managing workgroups can increase employee engagement, faster decision-making, and improve innovation.

Can Quality Circles coexist with other quality management initiatives?

Quality Circles can complement and coexist with other quality management methodologies like Six Sigma, each contributing to different improvement aspects.

Is Root Cause Analysis always necessary?

Root Cause Analysis is essential for resolving complex and recurring issues, but for simpler problems, it may be less extensive. However, a disciplined problem-solving approach is always recommended.

What is quality management tools?

Quality management tools are a set of techniques, methodologies, and instruments used to plan, control, and improve the quality of products and processes within an organization. These tools help businesses identify and rectify issues, optimize operations, and enhance overall performance. Quality management tools ensure that products and services meet or exceed customer expectations, fostering customer satisfaction and loyalty.

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The CIRCLES method: A PM’s guide to talking about design

A product manager’s responsibility is to identify the needs of their product’s users while also creating and improving the features that meets those needs. In doing so, PMs may sometimes rush straight to the solutions without fully understanding the problems these solutions will resolve. This can sometimes lead to needs that are half-met or solutions that are only partially solved.

Lewis C. Lin, an author and entrepreneur, created the CIRCLES method framework as a guideline to give complete and thoughtful responses to any design question. It is a good way to slow down the process and save PMs from jumping straight to the solution.

The product design questions are some of the most challenging to answer in a product management interview processes . The CIRCLES method provides a structure to solve these design questions accurately and ace your next PM interview.

What does CIRCLES stand for?

CIRCLES is an acronym that describes seven steps to solve a design problem. These linear steps are:

• Comprehend the situation
• Identify the customer
• Report the customer’s needs
• Cut though prioritization
• List solutions
• Evaluate trade-offs
• Summarize the recommendation

Let’s deep dive into each of these stages.

1. Comprehend the situation

This is the first step towards solving the design problem. The CIRCLES method suggests you should clarify as much information as possible.

You can ask the basic who, what, where, why and how questions. This allows you to comprehend the situation and grasp the context before moving further.

For example, you can ask:

• What is it?
• Who is it for?
• Where does it happen?
• Why do they need it?
• How does it work?

If the interviewer can’t clarify these questions, you should make assumptions based on what you know and convey those assumptions to the interviewer. This allows the interviewer to correct your assumptions if they think differently.

2. Identify the customer

The CIRCLES method guidelines focus on a single customer segment during the interview that lets you propose an amazing product. During the interview, the interviewer might not be able to connect with the segment you have established. Hence, a 2X2 matrix is a powerful way to visualize your persona that lists the archetypes’ behavior, demography, needs, goals, and frustration.

4. Report the customer’s needs

In this third step, the candidate describes the customer’s need, user requirements, or use cases. Typically it is presented in a popular format called user story to capture the user needs.

It typically follows this format:

“As a [ROLE] , I want [GOAL/DESIRE]  so that [OUTCOME/BENEFITS] .”

The point to remember is that the user story doesn’t describe how the solution works. Nevertheless, it is popular because user stories are concise, casual, and succinct.

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5. Cut through prioritization

Step four of the CIRCLES method is to cut through prioritization. This is simply the backlog item that a PM prioritizes based on value, time, and effort. Especially during interviews when a candidate doesn’t have enough time to walk through multiple user cases, picking one helps narrow down the choices and showcase the ability to prioritize and make decisions.

The prioritization criteria can be subjective, unlike real-time situations with a more quantitative approach. The prioritization matrix is a better way to articulate, and think through all the criteria. Typically, a candidate can use the matrix to decide on what is important, and each criterion is weighed over the other.

6. List solutions

Once the prioritization happens, it boils down to brainstorming the solutions. Here a PM comes up with as many ideas as possible, and can narrow them down at the next stage based on the feasibility, customer, and revenue criteria.

This is the stage where one gets the opportunity to raise creative solutions, and showcase the ability of lateral and innovative thinking. Besides, there are no perfect solutions, and this approach helps to complete this stage while taking care of other underlying facts.

7. Evaluate trade-offs

This stage is critical because the PM critiques their solution based on the trade-offs and risks. Unless the trade-off is unbiased, an impressive solution might come up. Finding the right solution also depends on all listed solutions’ short- and long-term pros and cons.

8. Summarize the recommendation

It is the final step when a PM candidate summarizes the final recommendation. The interviewer might expect the summary to be clear, concise, and eager to hear why the recommended product or feature benefits the user and cite reasons for preferring the solutions over others. The candidate will only do this stage if the interviewer asks for it; hence, it is optional. The trade-off steps already highlight the pros and cons of one solution over another.

How does the CIRCLES framework apply to product management?

Product managers need to keep the focus on the persona for which they are trying to solve problems. Besides, it is also important for them to communicate “why” they are building it. The CIRCLES method helps PMs prioritize features, feedback, experiments, and overall roadmap.

Secondly, the framework also allows PMs to ask the right questions during discovery to understand the actual pain points of the customers before jumping on the solutions. This overall approach helps PMs use lateral thinking rather than linear thinking while solving a problem.

Lateral thinking is an innovative way of building a product, whereas linear thinking is sequential and often jumps on the solutions quickly.

Alternative methods to the CIRCLES framework

The purpose of the CIRCLES framework is a great way to refrain oneself not to dive into the solution mode. However, explicitly using the CIRCLES method framework is not always necessary during interviews. There are other frameworks too that can be used as substitutes for the CIRCLES method.

Some of the more popular alternative methods are:

Double Diamond

Lean circle.

These methods, too, deal with defining the problem, prioritizing the problems, recommending solutions, picking the most coveted solutions (desirable, available, and feasible), and prototyping. All these substitutes embrace the concept of solving design problems without having a solution in mind. Let’s know about these alternative methods in detail.

The Double Diamond model is a popular framework developed by the British design council in 2005 based on converging and diverging thinking patterns. This framework is popularly used to solve and explore various problems and solutions before converging upon one.

The outcomes of this Double diamond are four stages:

Again, it is a non-linear thinking process that starts from the divergent thought process to explore the problems and root causes, which converge to define the problem. Further, the divergence approach is applied during the development stage to find the MVP by prioritizing high-value and low-effort solutions. Later, the solution is delivered and improved based on the feedback.

The Double Diamond approach is preferable in the workshop mode by designers as it is facilitator focused. In contrast, the CIRCLES method is best applied when attempting a design problem in an interview.

Double Diamond solves a design problem through detail prototyping, testing, validating the solution, and time-consuming process.

The Lean Circle is just a reduced version of the regular CIRCLES method. It reduces the seven stages to three simple stages:

• Explore the problem
• Define, prioritize and test the options
• Deliver a solution

It is especially used for innovative thinking and to avoid falling into the trap of linear thinking. A candidate should understand that it is not a hard and fast rule to follow the CIRCLES framework step by step during situations when time is a constraint. It shows the ability to communicate succinctly while demonstrating lateral thinking to avoid a one-size-fits-all approach.

Benefits of using the CIRCLES framework

In Decode and Conquer , Lewis C. Lin specifically mentioned that he developed the CIRCLES method to answer any design questions during product interviews. However, other than interviews, it can also be used in real-life situations and as an interim solution.

Some of the benefits of the CIRCLES framework are that it:

Allows for effective brainstorming

Identifies the best solutions, provides a structured answer.

PM can use the CIRCLES method to brainstorm an idea to solve a new feature problem. It also helps PMs communicate clearly and succinctly with the team on a new idea.

When budget and effort are hindrances to facilitating design sprints or Double Diamond workshops, a CIRCLES method can identify the best solutions the design team can work further to validate and test with users.

Of course, it is the best approach to quickly solve easy, complex, or unknown product design problems during product manager interviews. This framework provides the opportunity to clarify questions from the interviewer and then navigate through them.

Additionally, the CIRCLES method allows for structuring the answer and impresses the interviewer from beginning to end.

Weaknesses of the CIRCLES framework

The major drawback to the CIRCLES framework is that it is based on a few clarifying questions. The list of solutions comes up based on those clarifying questions, assumption of personas, and may be biased in prioritization.

Therefore, in practical situations, designers and PMs prefer using design-specific tools, like the Double Diamond.

Another fundamental issue with this method is that it narrows the scope of testing and validating the solution using actual prototypes. Instead, the solutions are evaluated based on the trade-offs of pre-conceived criteria, including customer experience, feasibility, and revenue.

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Circle Problems With Solutions

Clocks, wheels, dining plates, and coins. What do these things have in common?

They are all circular.

We are all familiar with what a circle looks like – it is round, has no sides, and has no start or end. Aside from identifying what it looks like, we also know how important it is to us. As early as the Mesopotamian civilization, people already used circular wheels to do their work and complete their journeys faster.

In this chapter, let us view circles from a “geometric lens” and explore how they’re helpful in our everyday lives. 

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Table of Contents

Definition of a circle.

In geometry, we formally define a circle as the set of all points that are equidistant (same distance) from a fixed point. That fixed point is called the center of the circle .

In the image above, point C is the center of the circle. If you take any two points in the circle, you are assured that these points have the same distance from point C. 

Note that when we mention the phrase “on the circle,” it means “in the frame of the circle.”

Suppose that points A and B are both in the circle. The distance from point C to point A is similar to that from point C to point B, or AC = BC.

Now, suppose that you take point D instead; the distance from point C to point D is certainly equal to the distance from point C to A and from point C to point B. Hence, AC = BC = CD.

Again, remember that all points on the circle have the same distance from the center.

Radius and Diameter of a Circle

The line segment formed by connecting the center to one of the points on the circle is called the radius.

If we draw a line segment from point C to point A in the figure above, we have formed a circle radius, which is segment AC since we have created a segment that connects the center (point C) and a point on the circle (point A).

Since all points on the circle are equidistant (same distance) from the center, all radii (plural of radius) have equal lengths. In other words, all radii are congruent segments. If someone tells you that one of the radii of a circle is longer than the other radius of the same circle, you know that it is false since all radii are equal in length. 

Meanwhile, if you draw a line segment from one point on a circle, pass through the center, and end on another point on the circle, then you have formed a diameter . Segment AQ below is an example of a diameter. The diameter of a circle is a line segment that divides the circle into two equal parts.

As you have noticed, a diameter consists of two radii (AC and CQ). The length of a circle’s diameter is equal to twice the length of a radius. 

d = 2r (where d is the diameter and r is the radius)

This also implies that the radius of a circle is ½ of the length of the diameter.

Also, like radii, all circle diameters are congruent or have the same length.

Circumference of a Circle

Suppose that you are watching a race in a circular field. How will you know the distance covered by the runners in that field?

The distance covered by the runners is equal to the total distance around the circular field. The total distance around a circle is called the circumference.

The circumference is the counterpart of the perimeter for circles. If squares, triangles, or rectangles have a perimeter as the total distance around them , the circumference is the total distance around a circle.

How do we determine the circumference of a circle?

We use the formula for the circumference of a circle which was derived as early as the ancient Greek civilization:

where C represents the circle’s circumference, d is the circle’s diameter, and π is a constant irrational number approximately equal to 3.1416. 

The number π (the Greek letter for “pi,” which is read as “pie”) is one of the most important numbers in mathematics since it’s used extensively not only in geometry but also in trigonometry and calculus . Many mathematicians provided an estimate for this irrational number, but one of the earliest estimations is from the Greek mathematician Archimedes.

π is the result when you try to divide the circumference of any circle on a flat surface by its diameter. It is an irrational number, meaning we cannot express it as a fraction with integers. One of the most common estimations of π used for calculation is 3.14 or 3.1416. However, these estimations are not the actual values of the number since the digits of π are never-ending. One of the most recent estimates of π has 62.8 trillion digits!

Going back to the circumference of a circle, the formula C = πd means that the distance around a circle is equal to the product of the irrational number and the circle’s diameter. Since the diameter of a circle is equivalent to twice the radius of a circle or d = 2r, we can rewrite the formula for the circumference of a circle as: 

where r is the radius of the circle.

This means that to calculate the circumference of a circle, you can use the following:

• Formula 1: C = πd (if the diameter is given)
• Formula 2: C = 2πr (if the radius is given)

Use formula 1 if the given in the problem is the circle’s diameter. However, if the radius is given, use formula 2.

Sample Problem 1: Determine the circumference of the circle below.

Solution: The circle above has a radius PQ with a length of 3 cm. Since the radius of the circle is given, let us use the formula C = 2πr.

C = 2π(3) since r = 3

Thus, the circumference of the circle is 6π cm.

Sometimes, the problem provides us with an estimated value of π that we should use in the problem. For instance, look at the given problems below.

Sample Problem 2: Compute the circumference of the circle below. Use π = 3.14.

Solution: The circle above has a radius of XY with a length of 5 cm. Since the radius of the circle is given, let’s use the formula C = 2πr.

C = 2(3.14)(5) Take note that the problem provides the value of π

Hence, the circumference of the circle is 31.4 cm.

Sample Problem 3: Calculate the circumference of a wheel with a diameter of 20 centimeters. Use π = 3.14.

Solution: We are given the diameter of the wheel. Thus, it is more convenient to use the formula C = πd.

C = (3.14)(20)

Hence, the circumference of the circle is 62.8 cm.

Sample Problem 4: The rim of a basketball ring has a circumference of 56 inches. Determine the diameter of the basketball ring.

Solution: The problem states that the rim of the basketball ring has a computed circumference of 56 inches. 

Recall that the circumference of a circle can be calculated using the formula:

We will use this formula instead of the other one since we are looking for the diameter of the ring.

The circumference is 56 inches. So, we have C = 56: 

The problem doesn’t provide a precise estimate of π, so we leave it as it is.

Let us now solve for the diameter (d) by dividing both sides by π:

πd = 56 Symmetric Property of Equality

πd⁄π = 56⁄π Dividing both sides of the equation by π

Thus, the diameter of the basketball ring is 56⁄π inches.

Sample Problem 5: The motorcycle’s wheel has a radius of 30 cm. How many revolutions must the wheel make to cover a total distance of 18.84 meters (1,884 centimeters)? Use π = 3.14.

Solution: When a wheel makes one revolution, it revolves at a total distance equivalent to its circumference. In other words, one revolution of the wheel = circumference of the circle .

Therefore, to find the number of revolutions it takes for the wheel to cover a total distance of 1,884 centimeters, we must divide 1,884 by the wheel’s circumference.

Let us first compute the circumference of the wheel.

The problem states that the wheel has a radius of 30 cm. Thus, we need to use the formula C = 2πr. Note that the problem requires us to use π = 3.14.

C = 2(3.14)(30) Substitute r = 30 and π = 3.14

Hence, the circumference of the wheel is 188.4 centimeters.

Now, to determine the number of revolutions the wheel will make to cover a total distance of 1,884 centimeters, let us divide 1,884 by the wheel’s circumference (188.4).

Number of revolutions = 1,884 ÷ 188.4 = 10

Thus, the answer to this problem is 10 revolutions .

Area of a Circle

The area of a circle refers to the amount of space a circle occupies on a flat surface. 

To calculate the area of a circle, we use the formula below.

A circle   = πr 2

where r is the circle’s radius, and π is the constant irrational number approximately equal to 3.1416.

The formula above states that the area of a circle is the product of the square of a circle’s radius and π.

How does the circumference differ from the area of a circle? Take a look at the image below.

The circumference is the total distance around a circle. It measures how long the boundary of that circle is. Meanwhile, the area measures how much space is occupied inside the circle.

Recall that when providing the area of a plane figure , we write the units in square unit form. For instance, if the circle’s radius is given in centimeters (cm), its area must be in square centimeters (cm 2 ). Similarly, if the unit of measurement is given in inches (in), then the area must be written in square inches (in 2 ). Take note of this reminder because even if your calculation is correct, but your unit is not, your final answer will be considered incorrect.

Sample Problem 1: A circle has a radius of 2 meters. Determine its area.

Solution: We only have to use the formula for the area of a circle to solve the problem. Furthermore, take note that the problem doesn’t provide an estimate of π. Thus, we use π as it is.

A circle = πr 2

A circle = π(2) 2 Substitute r = 2

A circle = 4π

Thus, the area of the circle is 4π m 2 .

Sample Problem 2: The radius of a circle is ½ inches long. Determine its area (Use π = 3.14).

Solution: Let’s use the formula for the area of a circle for this problem.

Please take note that we have r = ½ for this problem. However, converting ½ into a decimal form is much easier, so all of our figures in the formula will be decimal numbers. If you convert ½ into decimal form, you will get 0.5.

A circle = (3.14)(0.5) 2 Substitute r = 0.5

A circle = (3.14)(0.25)

A circle = 0.785

Therefore, the area of the circle is 0.785 square inches.

Sample Problem 3: Determine the area of the circle below (Use π = 3.14):

Solution: The image above provides us with the circle’s diameter of 10 cm. We cannot use the diameter to compute the area of the circle since the formula requires us to have the circle’s radius and not the diameter.

Recall that the diameter is equivalent to twice the radius of the circle. Hence, if the circle above has a diameter of 10 cm, then its radius is 10⁄2 = 5 cm.

Now that we have r = 5 cm. Let us compute the area of the circle using the formula:

A circle = (3.14)(5) 2 Substitute r = 5

A circle = (3.14)(25)

A circle = 78.5 

Thus, the area of the circle above is 78.5 cm 2 .

Sample Problem 4: The inscribed square (square inside the circle) in the image below has a perimeter of 24 cm. Meanwhile, the circle has a radius of 10 cm. Determine the area of the shaded region in the image below.

Solution: The problem above takes a little bit of analysis. How will you find the shaded region’s area in the image above?

If you try to remove the inscribed square in the image, you might notice that the remaining figure will be the shaded region. Hence, the area of the shaded region can be calculated if we are going to subtract the area of the square from the area of the circle:

A shaded region  = A circle  – A square

So, to find the area of the shaded region, we must first find the areas of the square and the circle.

Let us compute the area of the square first.

The problem states that the perimeter of the square is 24 cm. Recall that a square’s perimeter equals four times its side or P = 4s. Thus, if 24 is the perimeter of the square, then its side is:

24⁄4  = 4s⁄4 Dividing both sides of the equation by 4

Thus, the side of the square in the figure is 6 cm. We can now calculate the area of the square.

Recall the formula for the area of the square: A square = s 2

A square = s 2

A square = (6) 2 Since s = 6

A square = 36

Therefore the area of the square is 36 cm 2 .

Let us now compute the area of the circle.

The radius of the circle is 10 cm. Thus, we have r = 10. Let us use the formula for the area of a circle. Note that the problem does not provide us with an estimation of π that we should use, so we use π as it is.

A circle  = πr 2

A circle = π(10) 2

A circle = 100π

Hence, the area of the circle is 100π cm 2 .

We are not done yet. Remember that we are looking for the area of the shaded region. We have stated earlier that the area of the shaded region is equivalent to the difference between the area of the circle and the area of the square:

A shaded region  = A circle – A square

A shaded region = 100π – 36

That’s it! The area of the shaded region is 100π – 36 cm 2 .

Let us now move on to other concepts related to circles, such as the arcs and angles of a circle.

Arcs of a Circle

An arc is a portion of the circumference of a circle. It is formed by two points that are on the circle.

Measurement of an Arc

The measurement of an arc refers to the measure of the central angle of the circle that intercepts that arc. 

Let us go back to the previous example. You can create an angle that intercepts (or touches) the arc with the circle’s center as the vertex . That angle is called a central angle . 

In the figure above, we draw ∠XCY with C as the vertex. This is a central angle since its vertex is the circle’s center. Now, notice that this central angle intercepts or touches the arc XY.

The degree measurement of the central angle that touches the arc is the degree measurement also of that arc. Hence, in the figure above, if m∠XCY = 30°, then the measurement of the arc that it intercepts is also 30°.

Just like angles , we use degrees (°) as a unit of measurement for arcs.

The measurement of an arc can be any real number from 0° to 360°. The whole circle (the entire circumference) measures 360°.

We will discuss more central angles in the latter part of this review module, but in the meantime, this intuitive meaning of the measurement of an arc should be enough for you to continue to the next section.

Types of Arcs

We can classify arcs according to the degree measurement: minor arcs, semicircles, or major arcs.

1. Minor Arc

A minor arc is an arc with a measure that is less than 180°.  

Arc XY is an example of a minor arc since its degree measurement is only 30°, less than 180°.

2. Semicircle

A semicircle is an arc with a measure that is exactly 180°.

In the image above, arc PQ is a semicircle since it has a degree measurement of 180°. 

You might also realize that the semicircle is half of the circle. Thus, we can state that the distance of the arc created by a semicircle is equal to ½ of the circumference of the circle. Moreover, all circles have two semicircles.

3. Major Arc

A major arc has a degree measurement greater than 180° but less than 360°.

To name a major arc, we use three letters: two letters for the arc’s endpoints, and the third one is the point between these endpoints.

Arc ABC is a major arc since it has a measurement of 200°. 

The measurement of a major arc is equivalent also to 360 minus the measurement of the minor arc with the same endpoints as the major arc. 

For instance, if arc AC is 160°, we can compute the measurement of arc ABC by subtracting the 160° from 360°. Thus, the measurement of arc ABC is 200°.

Arc Addition Postulate

If an arc is formed by two adjacent arcs, then the measurement of the arc that is formed is equivalent to the sum of measurements of the adjacent arcs.

The arc addition postulate is analogous to the angle and segment addition postulates. 

In the figure above, arc AC is formed by adjacent arcs AB and BC. The arc addition postulate tells us that

Solution: By the arc addition postulate,

Chords, Secants, and Tangents of a Circle

This section discusses other circle parts, such as chords, secants, and tangents.

A chord is a segment with endpoints that are points on a circle . A secant is a line that intersects a circle in two points . Meanwhile, a tangent is a line intersecting a circle at exactly one point. 

In the figure above, segment AB is a chord since its endpoints are points on the circle (A and B). Meanwhile, Line l 1 is a secant since it is a line that intersects the circle at two points (which are C and D). Lastly, Line l ₂ is a tangent since it is a line that intersects the circle at exactly one point (which is at point Q). The point where the tangent and the circle intersect is called the point of tangency (Q is a point of tangency).

Also, note that the diameter is also a chord of a circle.

Some Theorems on Chords, Secants, and Tangents

Here are some important theorems that you must know about these three features of a circle:

1. Theorem: In a circle, two minor arcs are congruent if and only if their corresponding chords are congruent.

To better explain the theorem above, look at the given figure above.

PR and QS are congruent chords. This means that these chords have equal measurements or lengths. Since these chords are congruent, the theorem above tells us that their corresponding arcs, arc PQ and arc RS, are also congruent. 

2. Theorem: Tangent segments from a common external point are congruent.

In the figure above, line segments JK and JL have a common external point of J. As per the theorem above, since both of these segments are tangent to the circle (this means that each intersects the circle at exactly one point only), JK and JL are congruent or have equal lengths.

So suppose JK = 15 cm, then JL must be 15 cm since JK and JL are congruent according to the above theorem. 

3. Theorem (Segment of Chords Theorem): If two chords of a circle intersect, then the product of the segments of the first chord is equal to the product of the segments of the second chord.

To understand this theorem better, please refer to the image above. 

Chords AB and CD intersect at point R. The theorem tells us that the product of AB (AR and RB) segments is equal to the product of CD (CR and RD). In symbols: AR x RB = CR x RD.

Sample Problem 1: Find the value of x in the figure below.

Solution: In the figure above, x indicates the RB (or BR) measurement, a chord AB segment.

The Segment of Chords Theorem tells us that AR x RB = CR x RD (we have discussed this above). 

Let us use this theorem to find x:

AR ⋅ RB = CR ⋅ RD Replace the “x” sign with”⋅” since we are using x as a variable

12 ⋅ x = 10 ⋅ 6 Substituting the given values in the problem

12x = 60 Simplifying the equation

12x⁄12 = 60⁄12 Dividing both sides of the equation by 12

Hence, the value of x is 5. 

4. Theorem (Segment of Secants Theorem): If two secant segments have a common external endpoint, then the product of the lengths of the entire secant segment and its external segment is equal to the product of the lengths of the other entire secant segment and its external segment.

Let us take a look at the image above. Segments AC and AE are secant segments since both intersect two points of the circle. Furthermore, these segments share a common external endpoint which is A. Hence, the theorem above applies to these secant segments.

The theorem states that if we multiply the length of the entire secant segment AC by its external segment, which is AB (it is external since AB is outside the circle), then the result will be equal to the result when we multiply the entire segment AE by its external segment AD. 

In symbols: 

AC (entire secant segment) ⋅ AB(external segment of AC) = AE(entire secant segment) ⋅ AD (external segment of AE)

Sample Problem: Determine how long the segment PU is in the image below.

If you look carefully at the image above, PR and PU are secant segments that share a common external endpoint which is P. Thus, we can apply the Segment of Secants Theorem to find the length of segment PU. 

The Segment of the Secants Theorem allows us to conclude that the product of the entire secant segment PR and its external segment (which is PQ) is equal to the product of the entire secant segment PU and its external segment (which is PT).

Thus, we can write this equation:

PR ⋅ PQ = PU ⋅ PT

Let x be the length of segment PU:

PR ⋅ PQ = x ⋅ PT

Note that the segment PR is ten units long (since PR = PQ + QR by the segment addition postulate). 

Using the given values in the image above:

10 ⋅ 3  = x ⋅ 2

Simplifying the equation above:

By the symmetric property of equality

If we divide both sides of the equation by 2, we will obtain the following:

2x⁄2 = 30⁄2

Since x represents the length of the secant segment PU, then PU = 15 units.

Central Angles and Inscribed Angles

Angles can also be observed in a circle and have interesting properties you must know.

1. Central Angle

You have already encountered central angles in our earlier discussion about measuring an arc of a circle. Again, a central angle is an angle such that its vertex is the center of a circle, and its sides are the circle’s radii.

In the figure below, ABC is a central angle since its vertex is the circle’s center, B, and its sides are radii of the circle (AB and AC).

To reiterate what you have also learned earlier, the central angle and its intercepted arc have the same measurement. In other words, the central angle and its intercepted arc have the same measurement . Thus, in the figure above, ABC is congruent with its intercepted arc AC.

Sample Problem: Solve for the value of x below.

In the figure above, angle XYZ is a central angle that intercepts the arc XZ. We know that the measurement of the central angle is equal to its intercepted arc, thus:

x  + 5 = 30

We can now solve for x by simply transposing 5 to the right-hand side of the equation:

x  = -5 + 30

Thus, the value of x is 25.

2. Inscribed Angle

If the vertex of an angle is a point on the circle and its sides contain chords of the circle, then that angle is inscribed.

In the image above, FOR is an inscribed angle since its vertex is F, a point on the circle, and its sides, ray FO and FR, contain chords of the circles.

Inscribed Angle Theorem: “The measurement of an inscribed angle is equal to ½ of the measurement of its intercepted arc.”

The inscribed angle theorem provides us with a way to determine the measurement of an inscribed angle given its intercepted arc (or the arc it touches). According to the theorem, the measurement of an inscribed angle is just half of the measurement of its intercepted arc.

Thus, m∠FOR = 20°.

Using the inscribed angle theorem, can you prove that the angle will be right when you inscribe an angle in a semicircle?

Recall that a semicircle has a degree measurement of 180°. Thus, if you inscribe an angle in it, its measure will be half of the measurement of the semicircle. Half of 180° is 90°. Therefore, the inscribed angle is a right angle.

Hence, if an angle is inscribed in a semicircle, that angle is a right angle .

Inscribed Polygon

An inscribed polygon is one in which all vertices lie on a circle . Meanwhile, the circle that contains that polygon and touches its vertices is called a circumscribed circle.

In the image below, the trapezoid is an inscribed polygon, while the circle that contains it is a circumscribed circle.

If you inscribe a right triangle in a circle, the longest side of the right triangle (called the “hypotenuse”) will be the diameter of that circle. 

Next topic:  Right Triangles

Previous topic:  Volume of Solid Figures

Return to the main article:  The Ultimate Basic Math Reviewer

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in College Entrance Exam , LET , Reviewers , UPCAT

Last Updated June 10, 2023 12:29 AM

Jewel Kyle Fabula

Jewel Kyle Fabula is a Bachelor of Science in Economics student at the University of the Philippines Diliman. His passion for learning mathematics developed as he competed in some mathematics competitions during his Junior High School years. He loves cats, playing video games, and listening to music.

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