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SAT Math : Circles

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Example Questions

Example Question #31 : Circles

Circle a

The provided figure shows a circle on the coordinate axes with its center at the origin. $\overarc {AB}$ is a $45^{\circ }$ arc with length $\frac{3}{4} \pi$

Give the equation of the circle.

Possible Answers:

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

$x^{2}+ y^2 =9$

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

$x^{2}+ y^2 =3$

$x^{2}+ y^2 =6$

Correct answer:

$x^{2}+ y^2 =9$

Explanation:

A $45^{\circ }$ arc of a circle represents $\frac{45}{360 } = \frac{1}{8}$ of the circle, so the length of the arc is one-eighth its circumference - or, equivalently, the circumference is the length $\frac{3}{4} \pi$ multiplied by 8. Therefore, the circumference is

$C = \frac{3}{4} \pi \cdot 8 = 6 \pi$ .

The equation of a circle on the coordinate plane is

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

where (h, k) are the coordinates of the center and is the radius.

The radius of a circle can be determined by dividing its circumference by $2 \pi$ , so

$r = \frac{C}{2 \pi } = \frac{6 \pi }{2 \pi } = 3$

The center of the circle is (0,0) , so h = k = 0 . Substituting 0, 0, and 3 for , , and , respectively, the equation of the circle becomes

$(x- 0)^{2}+ (y - 0)^2 =3 ^{2}$ ,

$x^{2}+ y^2 =9$ .

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Example Question #32 : Circles

Circle a

The provided figure shows a circle on the coordinate axes with its center at the origin. The shaded region has area $20 \pi$ .

Give the equation of the circle.

Possible Answers:

$x^{2}+ y^2 =144$

$x^{2}+ y^2 =40$

$x^{2}+ y^2 =20$

$x^{2}+ y^2 =240$

$x^{2}+ y^2 =120$

Correct answer:

$x^{2}+ y^2 =120$

Explanation:

The shaded region, a $60^{\circ }$ sector of the circle, represents $\frac{60}{360} = \frac{1}{6}$ of the circle, so the area of the entire circle is six times the area of that region:

$A = 6 \cdot 20 \pi = 120 \pi$ .

The equation of a circle on the coordinate plane is

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

where (h, k) are the coordinates of the center and is the radius.

The formula for the area of a circle, given its radius , is

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

Set $A = 120 \pi$ and solve for $r^{2}$ :

$\pi r ^{2} = 120 \pi$

$\frac{\pi r ^{2} }{\pi}=\frac{ 120 \pi}{\pi}$

$r ^{2} = 120$

The center of the circle is (0,0) , so h = k = 0 . Substituting 0, 0, and 80 for , , and $r^{2}$ , respectively, the equation of the circle becomes

$(x- 0)^{2}+ (y - 0)^2 =120$

$x^{2}+ y^2 =120$

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Example Question #32 : How To Find The Equation Of A Circle

Circle a

The provided figure shows a circle on the coordinate axes with its center at the origin. The shaded region has an area of $20 \pi$ .

Give the equation of the circle.

Possible Answers:

$x^{2}+ y^2 =20$

$x^{2}+ y^2 =80$

Cannot be determined

$x^{2}+ y^2 =40$

$x^{2}+ y^2 =160$

Correct answer:

$x^{2}+ y^2 =80$

Explanation:

The shaded region represents one fourth of the circle, so the area of the entire circle is four times the area of that region:

$4 \cdot 20 \pi = 80 \pi$ .

The equation of a circle on the coordinate plane is

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

where (h, k) are the coordinates of the center and is the radius.

The formula for the area of a circle, given its radius , is

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

Set $A = 80 \pi$ and solve for $r^{2}$ :

$\pi r ^{2} = 80 \pi$

$\frac{\pi r ^{2} }{\pi}=\frac{ 80 \pi}{\pi}$

$r ^{2} = 80$

The center of the circle is (0,0) , so h = k = 0 . Substituting 0, 0, and 80 for , , and $r^{2}$ , respectively, the equation of the circle becomes

$(x- 0)^{2}+ (y - 0)^2 =80$ ,

$x^{2}+ y^2 =80$

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Example Question #153 : Coordinate Geometry

A circle is centered on point (5,12) . The area of the circle is $9\pi$ . What is the equation of the circle?

Possible Answers:

( x-5 )^2 + ( y -12 )^2 = 81

( x -12 )^2 + ( y- 5 )^2 = 9

( x -12 )^2 + ( y- 5 )^2 = 4.5

( x -5 )^2 - ( y - 12 )^2 = 9

( x- 5 )^2 + ( y -12 )^2 = 9

Correct answer:

( x- 5 )^2 + ( y -12 )^2 = 9

Explanation:

The formula for a circle is ( x -h )^2 + ( y -k )^2 = r^2

(h,k) is the coordinate of the center of the circle, therefore h = 5 and k = 12 .

The area of a circle:

$r^2\pi = 9\pi$

r^2 = 9

Therefore:

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

( x- 5 )^2 + ( y -12 )^2 = 9

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Example Question #2 : How To Find The Equation Of A Circle

A circle has a center at (5,5) and a radius of 2. If the format of the equation for the circle is (x-A)²+(y-B)²=C, what is C?

Possible Answers:

Correct answer:

Explanation:

The circle has a center at (5,5) and a radius of 2. Therefore, the equation is (x-5)²+(y-5)²=2², or (x-5)²+(y-5)²=4.

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SAT Math : Circles

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #31 : Circles

Example Question #32 : Circles

Example Question #32 : How To Find The Equation Of A Circle

Example Question #153 : Coordinate Geometry

Example Question #2 : How To Find The Equation Of A Circle

All SAT Math Resources

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