Question

what is the center of the circle given by the equation below?
$(x - 2)^{2}+(y + 8)^{2}=36$
a. $(-8,2)$
b. $(-2,8)$
c. $(2,8)$
d. $(2,-8)$

Explanation:

Step1: Recall the standard form of a circle's equation

The standard form of a circle's equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center of the circle.

Step2: Rewrite the given equation to match the standard form

Given \((x - 2)^2+(y+8)^2 = 36\), we can rewrite \((y + 8)^2\) as \((y-(-8))^2\).

Step3: Identify \(h\) and \(k\)

Comparing \((x - 2)^2+(y-(-8))^2 = 36\) with \((x - h)^2+(y - k)^2=r^2\), we get \(h = 2\) and \(k=-8\).

Answer:

D. \((2,-8)\)