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PSAT Math : How to find the equation of a circle

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

The endpoints of a diameter of circle A are located at points (-2,4) and (6,8) . What is the area of the circle?

Possible Answers:

$12\pi$

$80\pi$

$40\pi$

$160\pi$

$20\pi$

Correct answer:

$20\pi$

Explanation:

The formula for the area of a circle is given by A =πr². The problem gives us the endpoints of the diameter of the circle. Using the distance formula, we can find the length of the diameter. Then, because we know that the radius (r) is half the length of the diameter, we can find the length of r. Finally, we can use the formula A =πr² to find the area.

The distance formula is

The distance between the endpoints of the diameter of the circle is:

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

$d=\sqrt{8^{2}+4^{2}}$

$d=\sqrt{64+16}$

$d=\sqrt{80}=4\sqrt{5}$

To find the radius, we divide d (the length of the diameter) by two.

$r=\frac{d}{2}=\frac{4\sqrt{5}}{2}=2\sqrt{5}$

Then we substitute the value of r into the formula for the area of a circle.

$A=\pi r^{2}=\pi ({2\sqrt{5}})^{2}=20\pi$

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Example Question #231 : Coordinate Geometry

What is the equation for a circle of radius 9, centered at the intersection of the following two lines?

y_1 = 2x + 14

y_2 = 4x + 26

Possible Answers:

(x - 6)^2 + (y + 2)^2 = 81

(x + 3)^2 + (y - 1)^2 = 18

(x - 7)^2 + (y - 6.5)^2 = 9

(x + 6)^2 + (y - 2)^2 = 81

(x - 14)^2 + (y - 26)^2 = 9

Correct answer:

(x + 6)^2 + (y - 2)^2 = 81

Explanation:

To begin, let us determine the point of intersection of these two lines by setting the equations equal to each other:

2x + 14 = 4x + 26

14 = 2x + 26

-12 = 2x

x = -6

To find the y-coordinate, substitute into one of the equations. Let's use y_1 :

$y = 2 \cdot (-6) + 14$

y = -12 + 14

y = 2

The center of our circle is therefore (-6, 2) .

Now, recall that the general form for a circle with center at (x_0, y_0) is

(x - x_0)^2 + (y - y_0)^2 = r^2

For our data, this means that our equation is:

(x - (-6))^2 + (y - 2)^2 = 92

(x + 6)^2 + (y - 2)^2 = 81

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Example Question #31 : Circles

A circle is centered on point (5,12) . The area of the circle is $9\pi$ . What is the equation of the circle?

Possible Answers:

( x-5 )^2 + ( y -12 )^2 = 81

( x- 5 )^2 + ( y -12 )^2 = 9

( x -12 )^2 + ( y- 5 )^2 = 4.5

( x -12 )^2 + ( y- 5 )^2 = 9

( x -5 )^2 - ( y - 12 )^2 = 9

Correct answer:

( x- 5 )^2 + ( y -12 )^2 = 9

Explanation:

The formula for a circle is ( x -h )^2 + ( y -k )^2 = r^2

(h,k) is the coordinate of the center of the circle, therefore h = 5 and k = 12 .

The area of a circle:

$r^2\pi = 9\pi$

r^2 = 9

Therefore:

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

( x- 5 )^2 + ( y -12 )^2 = 9

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Example Question #233 : Coordinate Geometry

A circle with a radius of five is centered at the origin. A point on the circumference of the circle has an x-coordinate of two and a positive y-coordinate. What is the value of the y-coordinate?

Possible Answers:

$\sqrt{23}$

$\sqrt{21}$

Correct answer:

$\sqrt{21}$

Explanation:

Recall that the general form of the equation of a circle centered at the origin is:

x² + y² = r²

We know that the radius of our circle is five. Therefore, we know that the equation for our circle is:

x² + y² = 5²

x² + y² = 25

Now, the question asks for the positive y-coordinate when x = 2. To solve this, simply plug in for x:

2² + y² = 25

4 + y² = 25

y² = 21

y = ±√(21)

Since our answer will be positive, it must be √(21).

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PSAT Math : How to find the equation of a circle

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

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

Example Question #231 : Coordinate Geometry

Example Question #31 : Circles

Example Question #233 : Coordinate Geometry

All PSAT Math Resources

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