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Intermediate Geometry : Circles

Study concepts, example questions & explanations for Intermediate Geometry

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Example Questions

Example Question #1 : How To Find The Equation Of A Circle

What is the equation for a circle centered at (1,2) with a radius of ?

Possible Answers:

$(x-1)^{2}+ (y-2)^{2} = 9$

$(x-1)^{2}+ (y+2)^{2} = -9$

$(x-2)^{2}+ (y+1)^{2} = 18$

$(x+2)^{2}+ (y+1)^{2} = 36$

$(x+1)^{2}+ (y-2)^{2} = 9$

Correct answer:

$(x-1)^{2}+ (y-2)^{2} = 9$

Explanation:

The general equation for a circle centered at the origin is given by $x^{2}+y^{2}=r^{2}$ where is the radius of the circle.

To translate the origin to the first quadrant we need to subtract the appropriate amount to bring it back to center.

So the equation becomes $(x-1)^{2}+(y-2)^{2}=9$

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Example Question #1 : How To Find The Equation Of A Circle

A circle has a diameter starting at (3,2) and ending at (9,10) . What is the equation of the circle?

Possible Answers:

$(x-4)^{2}+(y-8)^{2}=25$

$(x+6)^{2}-(y+6)^{2}=25$

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

$(x-6)^{2}+(y-6)^{2}=25$

$(x-6)^{2}+(y-6)^{2}=100$

Correct answer:

$(x-6)^{2}+(y-6)^{2}=25$

Explanation:

Using the distance formula, the diameter is 10 units long, so the radius is 5. Then, by finding the midpoint of the diameter, you know the center of the cirle. Plugging values into the circle equation yields the final answer.

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

What is the equation of a circle with diameter , that is tangent to the x-axis at the point (3,0) and the y-axis at the point (0,3) ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The equation for a circle is (x-h)^2+(y-k)^2=r^2 where the center of the circle lies at the point and the radius of the circle is . If we are looking for a circle with a diameter of , then its radius must be . For the circle to be tangent to the x-axis at the point (3,0) and the y-axis at (0,3) , it must be centered at the point (3,3) . Therefore, the equation of the circle will be (x-3)^2+(y-3)^2=9 .

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Example Question #3 : How To Find The Equation Of A Circle

A circle has its center at the point (3,7) and a radius of units.

What is the equation of the circle?

Possible Answers:

(x-3)^2+(y-7)^2=16

(x-3)^2+(y-7)^2=4

(x+3)^2+(y+7)^2=16

x-3+y-7=16

Correct answer:

(x-3)^2+(y-7)^2=16

Explanation:

The equation for a circle is

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

where (h, k) is the center of the circle and r is the radius.

Plugging in the values of h=3 , k=7 , and r=4 , we get

(x-3)^2+(y-7)^2=16

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Example Question #4 : How To Find The Equation Of A Circle

Find the equation of a circle if the radius of the circle is and the center is located at the origin.

Possible Answers:

x^2+y^2 = 4

2x^2+y^2 = 1

$x^2+y^2 = \frac{1}{2}$

$x^2+y^2 = \sqrt2$

x^2+y^2 = 2

Correct answer:

x^2+y^2 = 4

Explanation:

The formula for the equation of a circle is:

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

The values of $\left ( h,k\right )$ represent the center, and both values are zero at the origin.

Plug in the known values and reduce.

(x-0)^2+(y-0)^2 = 2^2

x^2+y^2 = 4

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

Write the equation for a circle with the domain of [2,8] and a range of [-1,5] .

Possible Answers:

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

$(x-5)^{2}+(y-2)^{2}=9$

$(x+1)^{2}+(y-8)^{2}=36$

$(x-2)^{2}+(y-8)^{2}=9$

Correct answer:

$(x-5)^{2}+(y-2)^{2}=9$

Explanation:

The domain of a circle gives the width of the circle, the range of a circle gives the height of a circle - both of which equal the diameter.

The domain and range numbers are both 6 apart, meaning that the diameter of the circle is 6, which means the radius of the circle is 3.

The halfway point between the domain gives the x-coordinate for the center of the circle, which in this case is 5. 5 is halfway between 2 and 8.

The halfway point between the range gives the y-coordinate for the center of the circle, which in this case is 2. 2 is halfway between -1 and 5.

Now we have all of the information needed to plug into the equation for a circle in standard form as shown below where r is the radius and the center is (h,k):

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

Given a circle with radius of 3, and center of (5,2) we get the below equation of a circle.

$(x-5)^{2}+(y-2)^{2}=9$

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Example Question #3 : 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:

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

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

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

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

(x+1)^2 + (y-3)^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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Intermediate Geometry : Circles

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #1 : How To Find The Equation Of A Circle

Example Question #1 : How To Find The Equation Of A Circle

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

Example Question #3 : How To Find The Equation Of A Circle

Example Question #4 : How To Find The Equation Of A Circle

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

Example Question #3 : 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

All Intermediate Geometry Resources

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