Question

What is the center of the circle represented by this equation?
(x-7)2 + (y+4)2 =9

Answer 1

just a quick addition to the great reply above by "goddessboi"

`\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ (x-7)^2+(y+4)=9\implies [x-\stackrel{h}{7}]^2+[y-(\stackrel{k}{-4})]^2=3^2~\hfill \stackrel{vertex}{(7~~,~~-4)}`

Answer 2

Center is
`(7,-4)`

Explanation:

Recall that the equation for a circle is
`(x-h)^2+(y-k)^2=r^2` with center
`(h,k)` and radius
`r`. Hence, the center is
`(h,k)\rightarrow(7,-4)`.