How do you write an equation for a circle given center (8,-9) and passes through (21,22)?

1 Answer
Dec 12, 2016

Please see the explanation for steps leading to the equation.

Explanation:

The standard form for the equation of a circle is:

(x - h)^2 + (y - k)^2 = r^2" [1]"

where (x,y) is any point on the circle, (h,k) is the center point, and r is the radius.

We are given that the center point is (8, -9), therefore, we substitute 8 for h and -9 for k into equation [1]:

(x - 8)^2 + (y - -9)^2 = r^2" [2]"

NOTE: One can write (y - -9)^2 as (y + 9)^2 but doing so can cause errors when attempting to obtain the center point. Therefore, I do not recommend it.

To find the value of r, substitute the given point, 21 for x and 22 for y, into equation [2]:

(21 - 8)^2 + (22 - -9)^2 = r^2

(13)^2 + (31)^2 = r^2

169 + 961 = r^2

r^2 = 1130

r = sqrt(1130)

Substitute sqrt(1130) for r in equation [2]:

(x - 8)^2 + (y - -9)^2 = (sqrt(1130))^2" [3]"

Equation [3] is the standard form for the circle.