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 $r=$------.

Name: 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 $r=$------.
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The Answer of 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 $r=$------.

If the Circle x2+y2+12x−16y+51=0x^{2}+y^{2}+12 x-16 y+51=0x2+y2+12x−16y+51=0 Is Put into Standard Form (x−h)2+(y−k)2=r2(x-h)^{2}+(y-k)^{2}=r^{2}(x−h)2+(y−k)2=r2

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