jmc

Moving terms to the LHS, we have $x^2+4x+y^2-6y=-12$. Completing the square on the quadratic in $x$, we add $(4/2{)}^2=4$ to both sides. Completing the square on the quadratic in $y$, we add $(6/2{)}^2=9$ to both sides. We are left with the equation $x^2+4x+4+y^2-6y+9=1\Rightarrow (x+2{)}^2+(y-3{)}^2=1$. Thus, our circle has center $(-2,3)$. The distance between this center and the point $(1,7)$ is $\sqrt{(1-(-2){)}^2+(7-3{)}^2}=\boxed{5}$.