circle of problem solving

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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Circle Theorems

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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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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

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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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Chapter 1: Functions and Relations

Section 1.2: circles, learning outcomes.

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

DEFINITION OF A 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, [latex](h,k)[/latex], and the fixed distance is called the radius , [latex]r[/latex], of the circle.

DERIVING THE STANDARD FORM OF A CIRCLE

To derive the equation of a circle, we can use the distance formula with the points [latex](h,k)[/latex], [latex](x,y)[/latex], and the distance [latex]r[/latex].

Substitute the values.

Square both sides.

STANDARD FORM OF A CIRCLE

The standard form of a circle is as follows:

[latex]{\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}=r^{2}[/latex]

Write the Equation of a Circle in Standard Form

Example 1: write the standard form equation of a circle.

Write the standard form of a circle with radius [latex]3[/latex] and center [latex](0,0)[/latex].

Use the standard form of a circle.

Substitute in the values [latex]r=3, h=0, k=0[/latex].

  [latex]{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}=3^{2}[/latex]

[latex]{x}^{2}+{y}^{2}=9[/latex]

Example 2: WRITE THE STANDARD FORM equation OF A CIRCLE

Write the standard form of a circle with radius [latex]2[/latex] and center [latex](-1,3)[/latex].

Substitute in the values [latex]r=2, h=-1, k=3[/latex].

  [latex]{\left(x-(-1)\right)}^{2}+{\left(y-3\right)}^{2}=2^{2}[/latex]

[latex]{\left(x+1\right)}^{2}+{\left(y-3\right)}^{2}=4[/latex]

Example 3: finding the center and radius

Find the center and radius, then graph the circle: [latex]{\left(x+2\right)}^{2}+{\left(y-1\right)}^{2}=9[/latex].

Identify the center [latex](h,k)[/latex], and radius [latex]r[/latex].

  [latex]{\left(x-(-2)\right)}^{2}+{\left(y-1\right)}^{2}=3^{2}[/latex]

The center is [latex](-2,1)[/latex], and the radius is [latex]3[/latex].

Now graph the circle.  Plot the center first and then go up, down, left, and right 2 places.

Example 4: finding the center and radius

Find the center and radius, then graph the circle: [latex]{4x}^{2}+{4y}^{2}=64[/latex].

Divide each side by 4.

[latex]{x}^{2}+{y}^{2}=16[/latex]

  [latex]{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}=4^{2}[/latex]

The center is [latex](0,0)[/latex], and the radius is [latex]4[/latex].

Now graph the circle.  Plot the center first and then go up, down, left, and right 4 places.

GENERAL FORM OF A CIRCLE

The general form of a circle is as follows:

[latex]{x}^{2}+{y}^{2}+ax+by+c=0[/latex]

Example 5: WRITE THE STANDARD FORM Equation OF A CIRCLE

Find the center and radius, then graph: [latex]{x}^{2}+{y}^{2}-4x-6y+4=0[/latex].

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

[latex]{x}^{2}+{y}^{2}-4x-6y+4=0[/latex]

Group the x-terms and y-terms.  Collect the constants on the right right side.

  [latex]{x}^{2}-4x+{y}^{2}-6y=-4[/latex]

Complete the squares.

[latex]{x}^{2}-4x+4+{y}^{2}-6y+9=-4+4+9[/latex]

Rewrite as binomial squares.

[latex]{\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}=9[/latex]

The center is [latex](2,3)[/latex], and the radius is [latex]3[/latex].

Now graph the circle.  Plot the center first and then go up, down, left, and right 3 places.

Example 6: WRITE THE STANDARD FORM Equation OF A CIRCLE

Find the center and radius, then graph: [latex]{x}^{2}+{y}^{2}+8y=0[/latex].

[latex]{x}^{2}+{y}^{2}+8y=0[/latex]

Group the x-terms and y-terms.

  [latex]{x}^{2}+{y}^{2}+8y=0[/latex]

[latex]{x}^{2}+{y}^{2}+8y+16=0+16[/latex]

[latex]{\left(x-0\right)}^{2}+{\left(y+4\right)}^{2}=16[/latex]

The center is [latex](0,-4)[/latex], and the radius is [latex]4[/latex].

Example 7: APPLYING THE DISTANCE AND MIDPOINT FORMULAS TO A CIRCLE EQUATION

The diameter of a circle has endpoints [latex](-1,-4)[/latex] and [latex](7,2)[/latex]. Find the center and radius of the circle and also write its standard form equation.

The center of a circle is the center, or midpoint, of its diameter.  Thus the midpoint formula will yield the center point.

[latex]M=\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)[/latex]

  [latex]=\left(\frac{-1+7}{2},\frac{-4+2}{2}\right)[/latex]

[latex]=\left(\frac{6}{2},\frac{-2}{2}\right)[/latex]

[latex]=(3,-1)[/latex]

The center is [latex](3,-1)[/latex].  The distance formula will be used to find the distance from the center to one of the points on the circle.  This will yield the radius:

[latex]d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/latex]

[latex]d=\sqrt{(7-3)^{2}+(2-(-1))^{2}}[/latex]

[latex]d=\sqrt{4^{2}+3^{2}}=5[/latex]

The distance from the center to a point on the circle is 5.  Therefore the radius is 5.  The center and radius can now be used to find the standard form of the circle:

Start with the standard form of a circle.

Substitute in the values [latex]r=5, h=3, k=-1[/latex].

  [latex]{\left(x-3\right)}^{2}+{\left(y-(-1)\right)}^{2}=5^{2}[/latex]

[latex]{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}=25[/latex]

Example 8: Finding the Center of a Circle

The diameter of a circle has endpoints [latex]\left(-1,-4\right)[/latex] and [latex]\left(5,-4\right)[/latex]. Find the center of the circle.

The center of a circle is the center or midpoint of its diameter. Thus, the midpoint formula will yield the center point.

Key Concepts

• A circle is all points in a plane that are a fixed distance from a given point on the plane.  The given point is called the center, and the fixed distance is called the radius.
• The standard form of the equation of a circle with center [latex](h,k)[/latex] and radius [latex]r[/latex] is [latex]{\left(x-h\right)}^{2}{+}{\left(y-k\right)}^{2}=r^{2}[/latex]

Section 1.2 Homework Exercises

For the following exercises, write the standard form of the equation of the circle with the given radius and center [latex](0,0)[/latex].

1. Radius: [latex]7[/latex]

2. Radius: [latex]9[/latex]

3. Radius: [latex]\sqrt{2}[/latex]

4. Radius: [latex]\sqrt{5}[/latex]

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

5. Radius: [latex]1[/latex], center: [latex](3,5)[/latex]

6. Radius: [latex]10[/latex], center: [latex](-2,6)[/latex]

7. Radius: [latex]2.5[/latex], center: [latex](1.5,-3.5)[/latex]

8. Radius: [latex]1.5[/latex], center: [latex](-5.5,-6.5)[/latex]

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

9. Center: [latex](3,-2)[/latex] with point [latex](3,6)[/latex]

10. Center: [latex](6,-6)[/latex] with point [latex](2,-3)[/latex]

11. Center: [latex](4,4)[/latex] with point [latex](2,2)[/latex]

12. Center: [latex](-5,6)[/latex] with point [latex](-2,3)[/latex]

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

13. [latex]{\left(x+5\right)}^{2}+{\left(y+3\right)}^{2}=1[/latex]

14. [latex]{\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}=9[/latex]

15. [latex]{\left(x-4\right)}^{2}+{\left(y+2\right)}^{2}=16[/latex]

16. [latex]{\left(x+2\right)}^{2}+{\left(y-5\right)}^{2}=4[/latex]

17. [latex]x^{2}+{\left(y+2\right)}^{2}=25[/latex]

18. [latex]{\left(x-1\right)}^{2}+{y}^{2}=36[/latex]

19. [latex]{\left(x-1.5\right)}^{2}+{\left(y-2.5\right)}^{2}=0.25[/latex]

20. [latex]{\left(x-1\right)}^{2}+{\left(y-3\right)}^{2}=\frac{9}{4}[/latex]

21. [latex]x^{2}+y^{2}=64[/latex]

22. [latex]x^{2}+y^{2}=49[/latex]

23. [latex]2x^{2}+2y^{2}=8[/latex]

24. [latex]6x^{2}+6y^{2}=216[/latex]

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

25. [latex]x^{2}+y^{2}+2x+6y+9=0[/latex]

26. [latex]x^{2}+y^{2}-6x-8y=0[/latex]

27. [latex]x^{2}+y^{2}-4x+10y-7=0[/latex]

28. [latex]x^{2}+y^{2}+12x-14y+21=0[/latex]

29. [latex]x^{2}+y^{2}+6y+5=0[/latex]

30. [latex]x^{2}+y^{2}-10y=0[/latex]

31. [latex]x^{2}+y^{2}+4x=0[/latex]

32. [latex]x^{2}+y^{2}-14x+13=0[/latex]

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

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

35. 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.

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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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

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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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