problem: What is the distance between the center of the circle with equation $x^2+y^2=6x-8y+24$ and the point $(-3,-12)$?
solution: Moving terms to the LHS, we have $x^2-6x+y^2+8y=24$. Completing the square on the quadratic in $x$, we add $(6/2)^2=9$ to both sides. Completing the square on the quadratic in $y$, we add $(8/2)^2=16$ to both sides. We are left with the equation $x^2-6x+9+y^2+8y+16=49 \Rightarrow (x-3)^2+(y+4)^2=49$. Thus, our circle has center $(3,-4)$. The distance between this center and the point $(-3,-12)$ is $\sqrt{(-3-3)^2+(-12-(-4))^2}=\boxed{10}$.