GMAT Math : Circles

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Example Questions

Example Question #1 : Calculating The Equation Of A Circle

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

Possible Answers:

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

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

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

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

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

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

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

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Example Question #2 : 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:

$4\sqrt{2}\pi$

$10\pi$

$36\pi$

$100\pi$

$80\pi$

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:

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

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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

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

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Example Question #1 : Circles

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

Possible Answers:

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

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

center at (0, 0) and radius = 2

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

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

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Example Question #2 : Circles

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:

$10$

$16$

$32$

$5$

$25$

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 $a^{2}+b^{2}=r^{2}=5^{2}=25$ .

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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:

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

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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

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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

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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 $50 \pi - 25 \pi = 25 \pi$ .

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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

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

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

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GMAT Math : Circles

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Calculating The Equation Of A Circle

Example Question #2 : Calculating The Equation Of A Circle

Example Question #731 : Geometry

Example Question #1 : Circles

Example Question #2 : Circles

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

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