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Intermediate Geometry : Coordinate Geometry

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Example Questions

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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

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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

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

What is the equation of a circle that has its center at $(-\frac{1}{4}, 2)$ and a radius of ?

Possible Answers:

$(x+\frac{1}{4})^2+(y-2)^2=128$

$(x+\frac{1}{4})^2+(y-2)^2=256$

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

$(x+\frac{1}{4})^2+(y+2)^2=512$

Correct answer:

$(x+\frac{1}{4})^2+(y-2)^2=256$

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:

$(x+\frac{1}{4})^2+(y-2)^2=256$

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Example Question #22 : How To Find The Equation Of A Circle

What is the equation of a circle that has its center at (29, -100) and a radius of ?

Possible Answers:

(x-29)^2+(y-100)^2=256

$(x-29)^2+(y+100)^2=\sqrt{14}$

(x-100)^2+(y+29)^2=196

(x-29)^2+(y+100)^2=196

Correct answer:

(x-29)^2+(y+100)^2=196

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-29)^2+(y+100)^2=196

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Example Question #23 : How To Find The Equation Of A Circle

What is the equation of a circle that has its center at (45, -208) and a radius of $\frac{1}{4}$ ?

Possible Answers:

$(x-45)^2+(y+208)^2=\frac{1}{2}$

$(x-45)^2+(y+208)^2=\frac{1}{8}$

$(x-45)^2+(y+208)^2=\frac{1}{4}$

$(x-45)^2+(y+208)^2=\frac{1}{16}$

Correct answer:

$(x-45)^2+(y+208)^2=\frac{1}{16}$

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:

$(x-45)^2+(y+208)^2=\frac{1}{16}$

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Example Question #24 : How To Find The Equation Of A Circle

What is the equation of a circle that has its center at (-0.24, 12.4) and a radius of ?

Possible Answers:

(x-48)^2+(y-12.4)^2=25

(x+1.24)^2+(y-0.24^2=25

(x+0.24)^2+(y-12.4)^2=25

(x-0.24)^2-(y-12.4)^2=25

Correct answer:

(x+0.24)^2+(y-12.4)^2=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+0.24)^2+(y-12.4)^2=25

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Example Question #25 : How To Find The Equation Of A Circle

If the point (2,4) is on a circle whose center is at (6,8), which of the following is the standard form of the equation of this circle?

Possible Answers:

(x-6)^2+(y-8)^2=32

None of these.

(x+6)^2-(y+8)^2=16

(x-6)^2+(y-8)^2=16

(x+6)^2-(y+8)^2=32

Correct answer:

(x-6)^2+(y-8)^2=32

Explanation:

The standard equation of a circle is (x-h)^2+(y-k)^2=r^2 where (h,k) is the center and r is the radius. Since we were given a point on the circumference of the circle and its center, we use the distance formula to find the radius and then plug the radius and our center point into our equation.

$r=\sqrt{(6-2)^2+(8-4)^2}=\sqrt{32}$

Now substitute the center points and radius into the standard equation of a circle:

$(x-6)^2+(y-8)^2=(\sqrt{32})^2\rightarrow{(x-6)^2+(y-8)^2=32}$

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Example Question #26 : How To Find The Equation Of A Circle

Which of the following equations of a circle has the same center as the circle given by the equation x^2-8x+y^2-10y=-25 ?

Possible Answers:

(x-5)^2+(y-4)^2=1

(x+5)^2-(y-4)^2=16

(x+4)^2+(y-5)^2=20

(x-4)^2+(y-5)^2=10

Correct answer:

(x-4)^2+(y-5)^2=10

Explanation:

Recall the generic equation of a circle:

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

where the circle has a center at (h, k) and a radius of .

In order to find the center of the given circle, complete the squares with respect to both and .

x^2-8x+y^2-10y=-25

x^2-8x+16+y^2-10y+25=-25+16+25

(x-4)^2+(y-5)^2=16

Now, the circle with the equation (x-4)^2+(y-5)^2=10 is the only circle that also has its center at (4, 5) .

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Intermediate Geometry : Coordinate Geometry

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #17 : Circles

Example Question #18 : Circles

Example Question #19 : Circles

Example Question #20 : Circles

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

Example Question #22 : How To Find The Equation Of A Circle

Example Question #23 : How To Find The Equation Of A Circle

Example Question #24 : How To Find The Equation Of A Circle

Example Question #25 : How To Find The Equation Of A Circle

Example Question #26 : How To Find The Equation Of A Circle

All Intermediate Geometry Resources

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