jmc

We try to rewrite the given equation in the standard form for an ellipse. Completing the square in both variables, we have $$\begin{array}{rl}2(x^2+4x)+(y^2-10y)+c& =0\\ 2(x^2+4x+4)+(y^2-10y+25)+c& =33\\ 2(x+2{)}^2+(y-5{)}^2& =33-c.\end{array}$$To get this equation in standard form, we would normally try to divide by $33-c,$ and if $33-c>0,$ then we get the standard form of a (non-degenerate) ellipse. But we cannot do so if $33-c=0.$ Indeed, if $33-c=0,$ then only one point $(x,y)$ satisfies the equation, because both $x+2$ and $y+5$ must be zero for the left-hand side to equal zero. (And if $33-c<0$, then no points satisfy the equation, because the right-hand side is always nonnegative.) Thus, the value of $c$ that makes a degenerate ellipse satisfies $33-c=0,$ so $c=\boxed{33}.$