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

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

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)