If the Circle $x^{2}+y^{2}+12 x-16 y+51=0$ Is Put into Standard Form $(x-h)^{2}+(y-k)^{2}=r^{2}$

Short Answer

If the circle $x^{2}+y^{2}+12 x-16 y+51=0$ is put into standard form
$(x-h)^{2}+(y-k)^{2}=r^{2}$ , then $h=\ldots, k=\ldots$ , and $r=$ ------.