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SAT Math : Circles

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Example Question #21 : Circles

Circle a

The above figure shows a circle on the coordinate axes with its center at the origin. $\overarc {ACB}$ has length $35\pi$ .

Give the equation of the circle.

Possible Answers:

None of the other choices gives a correct response.

$x^{2}+ y^2 =900$

$x^{2}+ y^2 =42$

$x^{2}+ y^2 =30$

$x^{2}+ y^2 =1,764$

Correct answer:

$x^{2}+ y^2 =900$

Explanation:

$\overarc {A B}$ has measure $150 ^{\circ }$ , so $\overarc {ACB}$ , its corresponding major arc, measures $(360 - 150) ^{\circ } = 210 ^{\circ }$ , making it $\frac{210}{360 } = \frac{7}{12}$ of the circle. The length of $\overarc {ACB}$ , $21 \pi$ , is seven-twelfths its circumference, so set up the equation and solve for :

$\frac{7}{12} C = 35 \pi$

$\frac{12} {7} \cdot \frac{7}{12} C = \frac{12} {7} \cdot 35 \pi$

$C = 60\pi$

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

$r = \frac{C}{2 \pi } = \frac{60 \pi }{2 \pi } = 30$

The center of the circle is (0,0) , so h = k = 0 . Substituting 0, 0, and 30 for , , and , respectively, the equation of the circle becomes

$(x- 0)^{2}+ (y - 0)^2 =30^{2}$ ,

$x^{2}+ y^2 =900$ .

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Example Question #22 : Circles

Circle a

The above figure shows a circle on the coordinate axes with its center at the origin. $\overarc {AB}$ has length $6 \pi$ .

Give the equation of the circle.

Possible Answers:

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

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

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

$x^{2}+ y^2 =36$

$x^{2}+ y^2 =72$

Correct answer:

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

Explanation:

A $135^{\circ }$ arc of a circle represents $\frac{135}{360 } = \frac{3}{8}$ of the circle, so the length of the arc is three-eighths its circumference. Set up the equation and solve for :

$\frac{3}{8}C = 6 \pi$

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

$C = 16 \pi$

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

$r = \frac{C}{2 \pi } = \frac{16 \pi }{2 \pi } = 8$

The center of the circle is (0,0) , so h = k = 0 . Substituting 0, 0, and 8 for , , and , respectively, the equation of the circle becomes

$(x- 0)^{2}+ (y - 0)^2 =8^{2}$ ,

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

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Example Question #23 : Circles

Circle b

The above circle has area $36 \pi$ . Give its equation.

Possible Answers:

None of the other choices gives the correct response

(x - 16)^2+ y^2= 36

(x - 16)^2+ y^2= 256

$x ^{2}+ (y - 6)^2 = 36$

$x ^{2}+ (y - 16)^2 = 256$

Correct answer:

$x ^{2}+ (y - 6)^2 = 36$

Explanation:

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 area and the radius of a circle are related by the formula

$\pi r^{2} = A$

Set $A = 36\pi$ and solve for :

$\pi r^{2} = 36\pi$

$\frac{ \pi r^{2} }{\pi} = \frac{36\pi}{\pi}$

$r^{2} = 36$

$r = \sqrt{36} = 6$ .

The center of the circle lies on the -axis, so h = 0 . Also, the center is 6 units above the origin, so k = 6 . Setting h=0 ,k=6,r=6 , the equation becomes

$(x- 0)^{2}+ (y - 6)^2 = 6^{2}$

$x ^{2}+ (y - 6)^2 = 36$ .

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Example Question #24 : Circles

Circle b

The above circle has area $100 \pi$ . Give its equation.

Possible Answers:

$(x+50)^{2}+y^2 = 2,500$

$(x-50)^{2}+y^2 = 2,500$

$(x+10)^{2}+y^2 = 100$

$(x-10)^{2}+y^2 = 100$

None of the other choices gives the correct response

Correct answer:

$(x+10)^{2}+y^2 = 100$

Explanation:

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 area and the radius of a circle are related by the formula

$\pi r^{2} = A$

Set $A = 100\pi$ and solve for :

$\pi r^{2} = 100\pi$

$\frac{ \pi r^{2} }{\pi} = \frac{100\pi}{\pi}$

$r^{2} = 100$

$r = \sqrt{100} = 10$ .

The center of the circle lies on the -axis, so k = 0 . Also, the center is 10 unites left of the origin, so h = -10 . Setting h= -10,k=0,r =10 accordingly, the equation becomes

$(x- (-10 ))^{2}+ (y - 0)^2 = 10^{2}$

$(x+10)^{2}+y^2 = 100$ .

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Example Question #25 : Circles

Circle a

The above figure shows a circle on the coordinate axes with its center at the origin. The shaded region has area $49 \pi$ .

Give the equation of the circle.

Possible Answers:

$x^{2}+ y^2 =56$

$x^{2}+ y^2 =112$

$x^{2}+ y^2 =28$

$x^{2}+ y^2 =32$

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

Correct answer:

$x^{2}+ y^2 =56$

Explanation:

The unshaded region is a $45 ^{\circ }$ sector of the circle, making the shaded region a $(360 - 45) ^{\circ } = 305 ^{\circ }$ sector, which represents $\frac{305}{360} = \frac{7}{8}$ of the circle. Therefore, if is the area of the circle, the area of the sector is $\frac{7}{8}A$ . The sector has area $49 \pi$ , so

$\frac{7}{8}A = 49 \pi$

Solve for :

$\frac{8} {7} \cdot \frac{7}{8}A =\frac{8} {7} \cdot 49 \pi$

$A =56\pi$

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 $A = 56 \pi$ and solve for $r^{2}$ :

$\pi r ^{2} = 56 \pi$

$\frac{\pi r ^{2} }{\pi}=\frac{ 56 \pi}{\pi}$

$r ^{2} = 56$

The center of the circle is (0,0) , so h = k = 0 . Substituting 0, 0, and 56 for , , and $r^{2}$ , respectively, the equation of the circle becomes

$(x- 0)^{2}+ (y - 0)^2 =56$ ,

$x^{2}+ y^2 =56$ .

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Example Question #26 : Circles

Circle b

The above circle has circumference $32\pi$ . Give its equation.

Possible Answers:

(x - 16)^2+ y^2= 256

(x - 16)^2+ y^2= 32

$x ^{2}+ (y - 16)^2 = 32$

None of the other choices gives the correct response

$x ^{2}+ (y - 16)^2 = 256$

Correct answer:

$x ^{2}+ (y - 16)^2 = 256$

Explanation:

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, setting $C = 32\pi$ :

$r = \frac{C}{2 \pi } = \frac{32 \pi }{2 \pi } = 16$ .

The center of the circle lies on the positive -axis, so h = 0 . Also, the center is 16 units upward from the origin, so k = 16 . Setting h=0,k=r=16 , the equation becomes

$(x- 0)^{2}+ (y - 16)^2 = 16^{2}$

$x ^{2}+ (y - 16)^2 = 256$ .

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Example Question #27 : Circles

Circle b

The above circle has circumference $100 \pi$ . Give its equation.

Possible Answers:

None of the other choices gives the correct response

$(x+50)^{2}+y^2 = 2,500$

$(x-50)^{2}+y^2 = 2,500$

$x^{2}+(y+50)^2 = 2,500$

$x^{2}+(y-50)^2 = 2,500$

Correct answer:

$(x+50)^{2}+y^2 = 2,500$

Explanation:

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, setting $C = 100\pi$ :

$r = \frac{C}{2 \pi } = \frac{100 \pi }{2 \pi } = 50$ .

The center of the circle lies on the -axis, so k = 0 . Also, the center is 50 units left of the origin, so h= -50 . Setting h=0,k=-50,r=50 , the equation becomes

$(x- (-5 0))^{2}+y^2 = 50^{2}$

$(x+50)^{2}+y^2 = 2,500$ .

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Example Question #21 : How To Find The Equation Of A Circle

A circle is graphed on the coordinate plane. Its center is located at (3,6) and the circle intersects the x-axis at exactly one point. What is the equation defining this circle?

Possible Answers:

(x-6)^2+(y-3)^2=9

(x-3)^2+(y-6)^2=36

(x+3)^2+(y+6)^2=36

(x-3)^2+(y-6)^2=9

(x+6)^2+(y+3)^2=9

Correct answer:

(x-3)^2+(y-6)^2=36

Explanation:

Recall the general equation of a circle

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

The circle is understood to have a radius of length and to be centered at the point (h,k) .

The center (h,k) of the circle in the problem has been given as (3,6) . Hence, we can substitute the values h=3 and k=6 into the general equation of a circle to yield

(x-3)^2+(y-6)^2=r^2

We have also been given that the circle intersects the x-axis at exactly one point. By definition, the x-axis is tangent to the circle, and so the circle intersects the x-axis at the point (3,0) . This implies that the radius is the distance between this point on the x-axis and the center of the circle, and so we can calculate the radius using the distance formula, as shown:

$d=\sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$

$d=\sqrt{(3-3)^2+(0-6)^2}$

$d=\sqrt{36}=6$

Hence, the radius of this circle is units long and the equation of this circle is

(x-3)^2+(y-6)^2=6^2

(x-3)^2+(y-6)^2=36

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Example Question #21 : Circles

Circle a

The above circle has area $64 \pi$ . Give its equation.

Possible Answers:

$\textup{None of the other choices gives the correct response}$

$(x- 8)^{2}+ (y - 8)^2 = 8$

$(x- 8)^{2}+ (y - 8)^2 = 64$

$(x- 32)^{2}+ (y - 32)^2 = 1,024$

$(x- 32)^{2}+ (y - 32)^2 = 32$

Correct answer:

$(x- 8)^{2}+ (y - 8)^2 = 64$

Explanation:

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 area and the radius of a circle are related by the formula

$\pi r^{2} = A$

Set $A = 64 \pi$ and solve for :

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

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

$r^{2} = 64$

$r = \sqrt{64} = 8$

As seen below, the horizontal and vertical distance from the origin to the center of the circle are both equal to this radius, and it is located in Quadrant I, so the center is (8, 8) :

Circle b

Setting h = k = r = 8 , the equation of the circle becomes

$(x- 8)^{2}+ (y - 8)^2 = 8^{2}$

$(x- 8)^{2}+ (y - 8)^2 = 64$

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Example Question #21 : How To Find The Equation Of A Circle

Circle a

The provided circle has the following circumference:

$64 \pi$

Find its equation.

Possible Answers:

$(x- 8)^{2}+ (y - 8)^2 = 64$

$(x- 8)^{2}+ (y - 8)^2 = 8$

$(x- 32)^{2}+ (y - 32)^2 = 1,024$

None of these

$(x- 32)^{2}+ (y - 32)^2 = 32$

Correct answer:

$(x- 32)^{2}+ (y - 32)^2 = 1,024$

Explanation:

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, setting $C = 64 \pi$ :

$r = \frac{C}{2 \pi } = \frac{64 \pi }{2 \pi } = 32$ .

As can be seen in the diagram below, the center of the circle is the point in Quadrant I whose horizontal and vertical distance are both equal to the radius. Therefore, both coordinates are positive - and the center therefore has center at (32, 32) .

Circle b

Setting h = k = r = 32 , the equation of the circle becomes

$(x- 32)^{2}+ (y - 32)^2 = 32^{2}$

$(x- 32)^{2}+ (y - 32)^2 = 1,024$

Report an Error

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SAT Math : Circles

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #21 : Circles

Example Question #22 : Circles

Example Question #23 : Circles

Example Question #24 : Circles

Example Question #25 : Circles

Example Question #26 : Circles

Example Question #27 : Circles

Example Question #21 : How To Find The Equation Of A Circle

Example Question #21 : Circles

Example Question #21 : How To Find The Equation Of A Circle

All SAT Math Resources

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