jmc

As the center of the circle is at the point $(-5,2)$ and its radius is $r$, the equation for the circle is $(x+5{)}^2+(y-2{)}^2=r^2$. Expanding this, $$\begin{array}{rl}{x}^2+10x+25+y^2-4y+4& =r^2\\ x^2+y^2+10x-4y& =r^2-29.\end{array}$$Now, this equation must match the form $A{x}^2+2y^2+Bx+Cy=40,$ so we see that we can multiply the above by two so that the coefficients for $y^2$ match: $$2x^2+2y^2+20x-8y=2r^2-58.$$Thus, $A=2$, $B=20$, and $C=-8$. Also, $2r^2-58=40\Rightarrow 2r^2=98\Rightarrow r^2=49$. As $r$ is the radius, it must be positive, so $r=7$. Therefore, $A+B+C+r=2+20-8+7=\boxed{21}$.