jmc

Add the equations $y=(x+1{)}^2$ and $x+4=(y-3{)}^2$ to get $$x+y+4=(x+1{)}^2+(y-3{)}^2.$$(Any point that satisfies both equations must satisfy this equation as well.)

Completing the square in $x$ and $y$, we get $${\left(x+\frac{1}{2}\right)}^2+{\left(y-\frac{7}{2}\right)}^2=\frac{13}{2}.$$Thus, $r^2=\boxed{\frac{13}{2}}.$