GMAT Math : Circles

Study concepts, example questions & explanations for GMAT Math

Create An Account

All GMAT Math Resources

22 Diagnostic Tests 693 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #721 : Geometry

One of the diameters of a circle has endpoints (4, 5) and (10, 1). What is the equation of this circle?

Possible Answers:

$\small \small \small (x - 3)^{2} + (y - 7)^{2} = 13$

$\small \small (x - 7)^{2} + (y - 3)^{2} = 169$

$\small \small \small \small (x - 3)^{2} + (y - 7)^{2} = 169$

$\small 3x^{2} + 9y^{2} = 169$

$\small (x - 7)^{2} + (y - 3)^{2} = 13$

Correct answer:

$\small (x - 7)^{2} + (y - 3)^{2} = 13$

Explanation:

The equation of a circle with center (h,k) and radius is

$\small \small \small \small (x - h)^{2} + (y - k)^{2} = r^{2}$

The center is the midpoint of any diameter, so to find the center, we use the midpoint formula:

$\small \small h= \small \frac{x_{1}+x_{2}}{2}= \small \frac{4+10}{2} = 7$

$\small \small \small k= \small \frac{y_{1}+y_{2}}{2}= \small \frac{5+1}{2} = 3$

The center is (7,3). The radius is the distance between (7,3) and (10,1), so we use the distance formula:

$\small r = \sqrt{(10-7)^{2}+(1-3))^{2}} = \sqrt{3^{2}+(-2)^{2}} = \sqrt{13}$

So $\small r^{2} = 13$ , and the equation of the circle is

$\small (x - 7)^{2} + (y - 3)^{2} = 13$

Report an Error

Example Question #1 : Calculating The Equation Of A Circle

In the coordinate plane, a circle has center $\left ( 5,2 \right )$ and passes through the point $\left ( -1,-6 \right )$ . What is the area of the circle?

Possible Answers:

$100\pi$

$36\pi$

$80\pi$

$4\sqrt{2}\pi$

$10\pi$

Correct answer:

$100\pi$

Explanation:

The distance of the two points is $\sqrt{(5-(-1))^2+(2-(-6))^2}=10$ .

So 10 is the radius of the circle. Then we can calculate the area:

$A=\pi r^{2}=\pi \cdot 10^{2}=100\pi$ .

Report an Error

Example Question #731 : Geometry

A circle on the coordinate plane has area $64\pi$ ; its center is the origin. Which of the following is the equation of this circle?

Possible Answers:

$x ^{2} + y ^{2} = 64$

$x ^{2} + y ^{2} = 8$

$x ^{2} + y ^{2} = 4,096$

$x ^{2} + y ^{2} = 256$

$x ^{2} + y ^{2} = 16$

Correct answer:

$x ^{2} + y ^{2} = 64$

Explanation:

The equation of a circle with center at the origin is

$x ^{2} + y ^{2} = r^{2}$

where is the radius of the circle. The area of the circle is $A = \pi r ^{2}$ .

Since the area of the circle in the question is $64\pi$ , we can solve for $r ^{2}$ :

$A = \pi r ^{2} = 64 \pi$

$\pi r ^{2} \div \pi = 64 \pi \div \pi$

$r ^{2} = 64$

The equation is

$x ^{2} + y ^{2} = 64$

Report an Error

Example Question #1 : Calculating The Equation Of A Circle

Describe the circle given by the equation $(x+3)^{2}+y^{2} = 2$ .

Possible Answers:

center at (3, 0) and radius = $\sqrt{2}$

center at (0, -3) and radius = $\sqrt{2}$

center at (0, 3) and radius = 2

center at (0, 0) and radius = 2

center at (-3, 0) and radius = $\sqrt{2}$

Correct answer:

center at (-3, 0) and radius = $\sqrt{2}$

Explanation:

The equation for a circle is $(x-a)^{2}+(y-b)^{2} = r^{2}$ , where (a, b) is the center and r is the radius. In our equation, a = –3, b = 0, and r = $\sqrt{2}$ . Then the equation describes a circle with a center at (–3, 0) and a radius of $\sqrt{2}$ .

Report an Error

Example Question #2 : Calculating The Equation Of A Circle

In the -plane, point $(a,b)$ lies on a circle with center at the origin. The radius of the circle is 5. What is the value of $a^{2}+b^{2}$ ?

Possible Answers:

$25$

$5$

$10$

$32$

$16$

Correct answer:

$25$

Explanation:

$a$ and $b$ are the right-angle sides of a triangle, and the radius of the circle is the hypotenuse of the triangle. From the Pythagorean Theorem we would know that $a^{2}+b^{2}=r^{2}=5^{2}=25$ .

Report an Error

Example Question #6 : Calculating The Equation Of A Circle

$\textup{Circle }Q$ is centered at the point $\left ( -6,-5\right )$ and touches the y-axis once at the point $\left ( 0,-5\right )$ . What is the equation for $\textup{Circle }Q$ ?

Possible Answers:

$\small \small 6=(x-6)^2+(y-5)^2$

$\small 36=(x+6)^2+(y+5)^2$

$\small \small 6=(x+6)^2+(y+5)^2$

$\small \small 36=(x-6)^2+(y-5)^2$

$\small \small 25=(x+6)^2+(y+5)^2$

Correct answer:

$\small 36=(x+6)^2+(y+5)^2$

Explanation:

The equation of a circle in gneral form is:

$\small r^2=(x-h)^2+(y-k)^2$

In this case, our radius is 6, because our circle touches the y-axis once at the point (0,-5). This makes our radius equal to the absolute value of the x-coordinate of the center of our circle. Eliminate anything that doesn't have a 36.

Then, because our (h,k) are already negative, they change the signs to positive within the parentheses making our answer:

$\small 36=(x+6)^2+(y+5)^2$

Report an Error

Example Question #5 : Calculating The Equation Of A Circle

The points L (-4,5) and K (9,3) form a line which passes through the center of circle Q. Both points are on circle Q.

Which of the following represent the correct equation for circle Q?

Possible Answers:

6.58^2=(x+4)^2+(y+2.5)^2

6.58^2=(x-4)^2+(y-2.5)^2

13.15^2=(x+4)^2+(y+2.5)^2

6.58^2=(x-2.5)^2+(y-4)^2

6.58^2=(x+2.5)^2+(y+4)^2

Correct answer:

6.58^2=(x-2.5)^2+(y-4)^2

Explanation:

The general form for an equation of a circle is as follows:

r^2=(x-h)^2+(y-k)^2

Where r is the radius, and (h,k) are the coordinates of the center of the circle.

To begin, let's find the radius using distance formula. Because LK passes through the center of the circle and goes from the outer edge of the circle to the other side, we can say that LK is our diameter.

Use distance formula to find the length of LK.

$d=\sqrt{(x-x')^2+(y-y')^2}$

Plug in our points and simplify:

$d=\sqrt{(9--4)^2+(3-5)^2}=\sqrt{13^2+2^2}=\sqrt{173}\approx13.15$

That is our diameter, so our radius will be half of 13.15, or 6.575. This rounds to 6.58

Next, we can use midpoint formula to find the center of circle Q. Midpoint formula is:

$midpoint(x,y)=\left(\frac{x+x'}{2},\frac{y+y'}{2}\right)$

Plug in and simplify to find our midpoint

$midpoint(x,y)=\left(\frac{-4+9}{2},\frac{5+3}{2}\right)=\left(\frac{5}{2},\frac{8}{2}\right)=(2.5,4)$

Put it all together to get:

6.58^2=(x-2.5)^2+(y-4)^2

Report an Error

Example Question #5 : Calculating The Equation Of A Circle

Find the equation of a circle whose radius is and whose center is $\left (1,2 \right )$ .

Possible Answers:

(x-2)^2+(y-1)^2=16

(x-1)^2+(y-2)^2=4

(x-1)^2+(y-2)^2=16

(x-2)^2+(y-1)^2=4

Correct answer:

(x-1)^2+(y-2)^2=16

Explanation:

To solve this problem, remember that the general formula for a circle with center $\left ( h,k \right )$ and radius is:

(x-h)^2+(y-k)^2=r^2

Therefore,

(x-1)^2+(y-2)^2=16

Report an Error

Example Question #4 : Calculating The Equation Of A Circle

Two circles on the coordinate plane have the origin as their center. The outer circle has as its equation

$x^{2}+ y^2 = 50$ ;

the inner circle has as its equation

$x^{2}+ y^2 = 25$ .

Give the area of the region between the two circles.

Possible Answers:

$25 \pi$

$10 \pi$

$5 \pi$

Correct answer:

$25 \pi$

Explanation:

A circle with its center at the origin has as its equation

$x^{2}+ y^2 = r^{2}$ .

Let and be the radii of the larger and smaller circles, respectively.

The larger circle has equation $x^{2}+ y^2 = 50$ , so $R^{2} = 50$ . The area of a circle is equal to $\pi$ times the square of the radius, so the area of the larger circle is $\pi \cdot R^{2} = 50 \pi$ .

Similarly, the area of the smaller circle, whose equation is $x^{2}+ y^2 = 25$ , is $25 \pi$ .

The area of the region between them is the difference, which is $50 \pi - 25 \pi = 25 \pi$ .

Report an Error

Example Question #10 : Calculating The Equation Of A Circle

Two circles on the coordinate plane have the origin as their center. The inner circle has as its equation

$x^{2}+ y^2 = 64$ ;

the region between the circles has area $30 \pi$ .

Give the equation of the outer circle.

Possible Answers:

$x^{2}+ y^2 = 74$

$x^{2}+ y^2 =124$

$x^{2}+ y^2 = 90$

$x^{2}+ y^2 =94$

$x^{2}+ y^2 =79$

Correct answer:

$x^{2}+ y^2 =94$

Explanation:

A circle with its center at the origin has as its equation

$x^{2}+ y^2 = r^{2}$ .

Since the equation of the inner circle is $x^{2}+ y^2 = 64$ , then, for this circle,

$r^{2} = 64$ ,

and its area is $A = \pi r^{2} = \pi \cdot 64 = 64 \pi$ .

The area of the region between the circles is $30 \pi$ , so the outer circle has area

$64 \pi + 30 \pi = 94 \pi$ .

If is the radius of the outer circle, then its area is

$A = \pi R^{2} = 94\pi$

This makes $R ^{2} = 94$ , and the equation of the outer circle

$x^{2}+ y^2 = R^{2}$

$x^{2}+ y^2 =94$

Report an Error

All GMAT Math Resources

22 Diagnostic Tests 693 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Computer Science Tutors in San Diego, Math Tutors in San Diego, SAT Tutors in Atlanta, Reading Tutors in Boston, Calculus Tutors in Atlanta, ACT Tutors in Dallas Fort Worth, Biology Tutors in Miami, Physics Tutors in Chicago, Algebra Tutors in Philadelphia, MCAT Tutors in Dallas Fort Worth

Popular Courses & Classes

SAT Courses & Classes in Los Angeles, SAT Courses & Classes in Denver, Spanish Courses & Classes in Philadelphia, ACT Courses & Classes in San Francisco-Bay Area, SAT Courses & Classes in Seattle, GMAT Courses & Classes in Dallas Fort Worth, SAT Courses & Classes in Dallas Fort Worth, SAT Courses & Classes in San Diego, GMAT Courses & Classes in Los Angeles, LSAT Courses & Classes in New York City

Popular Test Prep

MCAT Test Prep in Houston, ISEE Test Prep in Washington DC, GMAT Test Prep in Seattle, MCAT Test Prep in Philadelphia, GRE Test Prep in Miami, GMAT Test Prep in Dallas Fort Worth, LSAT Test Prep in Philadelphia, GRE Test Prep in New York City, GMAT Test Prep in San Diego, MCAT Test Prep in Phoenix

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

Email address:
Your name:
Feedback:

GMAT Math : Circles

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #721 : Geometry

Example Question #1 : Calculating The Equation Of A Circle

Example Question #731 : Geometry

Example Question #1 : Calculating The Equation Of A Circle

Example Question #2 : Calculating The Equation Of A Circle

Example Question #6 : Calculating The Equation Of A Circle

Example Question #5 : Calculating The Equation Of A Circle

Example Question #5 : Calculating The Equation Of A Circle

Example Question #4 : Calculating The Equation Of A Circle

Example Question #10 : Calculating The Equation Of A Circle

All GMAT Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details