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Intermediate Geometry : Coordinate Geometry

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Example Questions

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Example Question #7 : How To Find The Equation Of A Circle

Write the equation for a circle with center (2,5) passing through the point (-1, 3)

Possible Answers:

(x+1)^2 + (y-3)^2 = 13

$(x-1)^2 + (y-3)^2 = \sqrt{5}$

$(x+2)^2 + (y+5)^2 = \sqrt{13}$

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

(x-2)^2 + (y-5)^2 = 13

Correct answer:

(x-2)^2 + (y-5)^2 = 13

Explanation:

To determine the equation for a circle, we need to know the center and the radius. In this case, we know the center and one of the points on the circle. The radius of the circle is the distance from the center to this point, so to determine it, use the distance formula:

$r = \sqrt{(2+1)^2 + (5-3)^2} = \sqrt{9+4} = \sqrt{13}$

The equation for a circle is written as (x-h)^2 + (y-k)^2 = r^2 where the center is (h,k) and the radius is r. For this circle, plug in (2,5) and $r = \sqrt{13}$ :

$(x-2)^2 + (y-5)^2 = \sqrt{13}^2$ or more simply

(x-2)^2 + (y-5)^2 = 13

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Example Question #8 : How To Find The Equation Of A Circle

Write the equation for a circle passing through the point (-1, 3) and centered at the origin.

Possible Answers:

$(x+1)^2 + (y-3)^2 = \sqrt{8}$

$x^2 + y^2 = \sqrt{10}$

x^2 + y^2 = 10

x^2 + y^2 = 8

(x+1)^2 + (y-3)^2 = 10

Correct answer:

x^2 + y^2 = 10

Explanation:

To determine the equation of the circle, we need to know the radius, or the distance from the origin to the point on the circle. Use the distance formula:

$r = \sqrt{(-1-0)^2 + (3-0)^2 } = \sqrt{10}$

Since the circle is centered at the origin, its equation is $(x-0)^2 + (y-0)^2 = \sqrt{10}^2$ or more simply x^2 + y^2 = 10

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Example Question #9 : How To Find The Equation Of A Circle

Which of these points is inside the circle

(x-2)^2 + (y-1)^2 = 10 ?

Possible Answers:

(1, 4)

(1,8)

(10, 0)

(1, 3)

(3, -2)

Correct answer:

(1, 3)

Explanation:

Plugging in the point (1, 3) shows that this point is inside the circle, since the left side of the equation will be less than the right:

(1-2)^2+(3-1)^2 = 10

(-1)^2 + (2)^2 = 10

1+4 = 5 < 10

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Example Question #10 : How To Find The Equation Of A Circle

Write the equation for the circle with center (5, 4) and radius

Possible Answers:

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

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

(x+5)^2 + (y+4)^2 = 400

(x-5)^2 + (y-4)^2 = 400

(x-5)^2 + (y-4)^2 = 100

Correct answer:

(x-5)^2 + (y-4)^2 = 400

Explanation:

The equation for a circle is in the form (x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius.

In this case:

(x-5)^2 + (y-4)^2 = 20^2 = 400

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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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Intermediate Geometry : Coordinate Geometry

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #7 : How To Find The Equation Of A Circle

Example Question #8 : How To Find The Equation Of A Circle

Example Question #9 : How To Find The Equation Of A Circle

Example Question #10 : How To Find The Equation Of A Circle

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

All Intermediate Geometry Resources

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