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Intermediate Geometry : How to find the equation of a circle

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Example Question #1601 : Intermediate Geometry

What is the equation for a circle centered at (8, 0) passing through the point (-2, -3)

Possible Answers:

(x+2)^2 + (y+3)^2 = 109

(x-8)^2 + y = 91

x^2 + (y-8)^2 = 109

(x+2)^2 + (y+3)^2 = 91

(x-8)^2 + y = 109

Correct answer:

(x-8)^2 + y = 109

Explanation:

The equation of a circle is defined by (x-h)^2 + (y-k)^2 = r^2 where the center is (h,k) and the radius is r. The radius is the distance from the center to any point on the circle, so we can use the distance formula to calculate it:

$r= \sqrt{(8+2)^2 + (0+3)^2} = \sqrt{100+9} = \sqrt{109}$

The equation is then $(x-8)^2 + (y-0)^2 = \sqrt{109}^2$ or more simplified, (x-8)^2 + y^2 = 109

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Example Question #12 : Circles

Write the equation for a circle with center (1,-3) and passing through the point (0,4)

Possible Answers:

(x+1)^2 + (y-3)^2 = 48

(x-1)^2 + (y+3)^2 = 48

(x-1)^2 + (y+3)^2 = 50

(x-1)^2 + (y-3)^2 = 2

(x+1)^2 + (y-3)^2 = 2

Correct answer:

(x-1)^2 + (y+3)^2 = 50

Explanation:

To determine the radius of the circle, use the distance formula:

$r = \sqrt{(1-0)^2 + (-3-4)^2} = \sqrt{1+49} = \sqrt{50}$

The equation for the circle is $(x-1)^2 + (y--3)^2 = \sqrt{50}^2$ or more simply (x-1)^2 + (y+3)^2 = 50

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Example Question #11 : Circles

What is the equation of a circle that has its center at (-10, 2) and a radius of ?

Possible Answers:

(x+10)^2+(y-2)=144

(x+10)^2+(y-2)=24

(x+10)^2+(y+2)=144

(x-10)^2+(y-2)=36

Correct answer:

(x+10)^2+(y-2)=144

Explanation:

Recall the standard form for the equation of a circle:

(x-a)^2+(y-b)^2=r^2

In this equation, (a, b) represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

(x+10)^2+(y-2)=144

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Example Question #14 : Circles

What is the equation of a circle that has its center at (-2, 14) and a radius of ?

Possible Answers:

(x-2)^2+(y+14)^2=16

(x-14)^2+(y-2)^2=4

(x+2)^2+(y-14)^2=16

(x-2)^2-(y+14)^2=4

Correct answer:

(x+2)^2+(y-14)^2=16

Explanation:

Recall the standard form for the equation of a circle:

(x-a)^2+(y-b)^2=r^2

In this equation, (a, b) represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

(x+2)^2+(y-14)^2=16

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Example Question #15 : Circles

What is the equation of a circle that has its center at (12, 15) and a radius of ?

Possible Answers:

(x-12)^2-(y+15)^2=16

(x-12)^2+(y-15)^2=8

(x-12)^2+(y-15)^2=16

(x+12)^2+(y+15)^2=16

Correct answer:

(x-12)^2+(y-15)^2=16

Explanation:

Recall the standard form for the equation of a circle:

(x-a)^2+(y-b)^2=r^2

In this equation, (a, b) represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

(x-12)^2+(y-15)^2=16

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Example Question #16 : Circles

What is the equation of a circle that has its center at $(5, \frac{1}{4})$ and a radius length of $\frac{1}{5}$ ?

Possible Answers:

$(x-5)^2-(y+\frac{1}{4})^2=\frac{1}{5}$

$(x-5)^2+(y-\frac{1}{4})^2=\frac{1}{25}$

$(x-5)^2+(y-\frac{1}{4})^2=\frac{1}{10}$

$(x-\frac{1}{4})^2+(y+5)^2=\frac{1}{25}$

Correct answer:

$(x-5)^2+(y-\frac{1}{4})^2=\frac{1}{25}$

Explanation:

Recall the standard form for the equation of a circle:

(x-a)^2+(y-b)^2=r^2

In this equation, (a, b) represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

$(x-5)^2+(y-\frac{1}{4})^2=\frac{1}{25}$

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Example Question #17 : Circles

What is the equation of a circle that has its center at (15, 2) and a radius of $\frac{3}{5}$ ?

Possible Answers:

$(x-15)^2+(y-2)^2=\frac{3}{5}$

$(x-15)^2+(y-2)^2=\frac{9}{25}$

$(x-15)^2+(y+2)^2=\frac{6}{5}$

$(x+15)^2+(y+2)^2=\frac{9}{25}$

Correct answer:

$(x-15)^2+(y-2)^2=\frac{9}{25}$

Explanation:

Recall the standard form for the equation of a circle:

(x-a)^2+(y-b)^2=r^2

In this equation, (a, b) represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

$(x-15)^2+(y-2)^2=\frac{9}{25}$

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Example Question #18 : Circles

What is the equation of a circle that has its center of (124, -201) and a radius of ?

Possible Answers:

(x-201)^2+(y+124)^2=225

(x-124)^2+(y+201)^2=225

(x-124)^2+(y+201)^2=15

$(x-124)^2+(y-201)^2=\sqrt{15}$

Correct answer:

(x-124)^2+(y+201)^2=225

Explanation:

Recall the standard form for the equation of a circle:

(x-a)^2+(y-b)^2=r^2

In this equation, (a, b) represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

(x-124)^2+(y+201)^2=225

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Example Question #19 : Circles

What is the equation of a circle that has its center of (19, 28) and a radius of ?

Possible Answers:

$(x-19)^2+(y-28)^2=\sqrt{13}$

(x-19)^2+(y-28)^2=169

$(x-19)^2+(y-28)^2=\frac{13}{2}$

(x-19)^2+(y-28)^2=26

Correct answer:

(x-19)^2+(y-28)^2=169

Explanation:

Recall the standard form for the equation of a circle:

(x-a)^2+(y-b)^2=r^2

In this equation, (a, b) represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

(x-19)^2+(y-28)^2=169

Report an Error

Example Question #20 : Circles

What is the equation of a circle that has its center at (-22, -120) and a radius of ?

Possible Answers:

(x+120)^2+(y+22)^2=40

$(x+22)^2+(y+120)^2=2\sqrt5$

(x-22)^2-(y+120)^2=100

(x+22)^2+(y+120)^2=400

Correct answer:

(x+22)^2+(y+120)^2=400

Explanation:

Recall the standard form for the equation of a circle:

(x-a)^2+(y-b)^2=r^2

In this equation, (a, b) represents the center of the circle and is the radius of the circle.

Since the center and radius of the circle are given to us, just substitute in the information itno the standard form of the equation of the circle.

The equation of the given circle is:

(x-19)^2+(y-28)^2=169

Report an Error

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Intermediate Geometry : How to find the equation of a circle

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #1601 : Intermediate Geometry

Example Question #12 : Circles

Example Question #11 : Circles

Example Question #14 : Circles

Example Question #15 : Circles

Example Question #16 : Circles

Example Question #17 : Circles

Example Question #18 : Circles

Example Question #19 : Circles

Example Question #20 : Circles

All Intermediate Geometry Resources

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