jmc

We will complete the square to determine the standard form equation of the circle. Shifting all but the constant term from the RHS to the LHS, we have $x^2-8x+y^2+6y=-20$. Completing the square in $x$, we add $(-8/2{)}^2=16$ to both sides. Completing the square in $y$, we add $(6/2{)}^2=9$ to both sides. The equation becomes $$\begin{array}{rl}{x}^2-8x+y^2+6y& =-20\\ \Rightarrow x^2-8x+16+y^2+6y+9& =5\\ \Rightarrow (x-4{)}^2+(y+3{)}^2& =5\end{array}$$ Thus, the center of the circle is at point $(4,-3)$ so $x+y=4+(-3)=\boxed{1}$.