x^{2}+(y - 8)^{2}=0 give the center

$x^{2}+(y - 8)^{2}=0 give the center$

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

Given $x^{2}+(y - 8)^{2}=0$. Here $h = 0$ (since $x-0=x$) and $k = 8$.

$(0,8)$