Solution:

Completing the square, $2y = (x - 4)^2 - 4,$ so with $u = x - 4$ the parabola is the set of points $\left(4 + u,\ \frac{u^2}{2} - 2\right)$ and the center $(4, 39)$ lies on its axis. The circle is tangent to the parabola at a point exactly when the two curves share a tangent line there, i.e. when the radius to that point is normal to the parabola — which happens exactly at critical points of the squared distance $$D(u) = u^2 + \left(\frac{u^2}{2} - 41\right)^2, \qquad D'(u) = 2u + u\left(u^2 - 82\right) = u\left(u^2 - 80\right).$$

At $u = \pm\sqrt{80}:$ $D = 80 + (40 - 41)^2 = 81,$ so $r = 9$ (the circle touches the parabola at two symmetric points). At $u = 0,$ the point is the vertex $(4, -2)$ at distance $41,$ where the parabola and the circle of radius $41$ both have horizontal tangent lines, so $r = 41$ also works.

The sum is $9 + 41 = 50.$