jmc

We complete the square for the first equation by observing that the first equation is equivalent to $$(x^2-10x+25)+(y^2-4y+4)=36,$$ which is also equivalent to $$(x-5{)}^2+(y-2{)}^2=6^2.$$ Similarly, the equation for the second circle is $$(x+7{)}^2+(y+3{)}^2=3^2.$$ Hence, the centers of the circles are $(5,2)$ and $(-7,-3)$, and the radii of the circles are equal to 6 and 3, respectively. The distance between the points $(5,2)$ and $(-7,-3)$ by the distance formula is $\sqrt{(5-(-7){)}^2+(2-(-3){)}^2}=\sqrt{12^2+5^2}=\sqrt{169}=13$. Therefore, to find the shortest distance between the two circles, we must subtract from $13$ the sum of the radii of the two circles. Thus, the shortest distance between the circles is $13-3-6=\boxed{4}$.