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 $x^2+2x+y^2+10y=-16$. Completing the square in $x$, we add $(2/2{)}^2=1$ to both sides. Completing the square in $y$, we add $(10/2{)}^2=25$ to both sides. The equation becomes $$\begin{array}{rl}{x}^2+2x+y^2+10y& =-16\\ \Rightarrow x^2+2x+1+y^2+10y+25& =10\\ \Rightarrow (x+1{)}^2+(y+5{)}^2& =10\end{array}$$ Thus, the center of the circle is at point $(-1,-5)$ so $x+y=-1+(-5)=\boxed{-6}$.