To find the center of the circle, I need to complete the square for both $x$ and $y$ terms. I can rewrite the equation as $(x^2+2x)+(y^2+10y)=-16$. To complete the square for $(x^2+2x)$, I need to add $(2/2)^2=1$ to both sides. To complete the square for $(y^2+10y)$, I need to add $(10/2)^2=25$ to both sides. So I have $(x^2+2x+1)+(y^2+10y+25)=-16+1+25$. Simplifying, I get $(x+1)^2+(y+5)^2=10$. Now I can see that the center of the circle is $(-1,-5)$, since the equation is in the form $(x-h)^2+(y-k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius. Therefore, $x+y=(-1)+(-5)=-6$. 
The center of the circle with equation $x^2+y^2=-2x-10y-16$ is the point $(x,y)$. What is $x+y$?