jmc

We complete the square for the first equation by adding $(-24/2{)}^2$ and $(-32/2{)}^2$ to both sides, which gives $$(x^2-24x+144)+(y^2-32y+256)-16=0,$$ which is also equivalent to $$(x-12{)}^2+(y-16{)}^2=4^2.$$ Similarly, the equation for the second circle is $$(x+12{)}^2+(y+16{)}^2=4^2.$$ Hence, the centers of the circles are $(12,16)$ and $(-12,-16)$ respectively. Furthermore, the radii of the circles are equal to $4$. Now the distance between the points $(12,16)$ and $(-12,-16)$ by the distance formula or similarity of $3-4-5$ triangles is $40$. Therefore, to find the shortest distance between the two circles, we must subtract from $40$ the distances from the centers to the circles. Thus, the shortest distance between the circles is $40-4-4=\boxed{32}$.