If the equation of the circle is $x^2+y^2-8x+10y-12=0$, then 

I. Centre of the circle is ( 4, -5).

II. Radius of the circle is $\sqrt{53}$.

The given equation is

$\begin{array}{l}{x}^2+y^2-8x+10y-12=0\\ \left(x^2-8x\right)+\left(y^2+10\right)=12\\ \left(x^2-8x+16\right)+\left(y^2+10y+25\right)=12+16+25\\ (x-4{)}^2+(y+5{)}^2=53\end{array}$

Therefore, the given circle has centre at (4, -5) and radius $\sqrt{53}$.