jmc

If we complete the square after bringing the $x$ and $y$ terms to the other side, we get $$(x-5{)}^2+(y+3{)}^2=0.$$Squares of real numbers are nonnegative, so we need both $(x-5{)}^2$ and $(y+3{)}^2$ to be $0.$ This only happens when $x=5$ and $y=-3.$ Thus, $x+y=5+(-3)=\boxed{2}.$