Question

identify the center of the circle whose equation is (x-2)^2+(y+8)^2=16

Answer 1

The equation of a circle is

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

So this equation you provided has:
 Center: (2,-8)
 Radius: 4

Answer 2

Answer:

The center of the circle is (2, -8)

Step-by-step explanation:

Given : the equation of circle as $(x-2)^2+(y+8)^2=16$

We have to find the center of the circle whose equation is  $(x-2)^2+(y+8)^2=16$

The general equation of a circle is given as $(x-h)^2+(y-k)^2=r^2$

Where , (h,k) represents the center of the circle and r represents the radius of the circle.

Comparing the given equation of circle  $(x-2)^2+(y+8)^2=16$ with the general we get,

h = 2 , k = -8 and r = 4

Thus  $(x-2)^2+(y+8)^2=16$ can be written as $(x-2)^2+(y-(-8))^2=4^2$

Thus, the center of the circle is (2, -8)