If (-6,0), (0, 6), and (-7,7) are the vertices of a $∆$ABC. The incircle of the triangle has equation

The triangle is evidently isosceles and, therefore, the median through C is the angle bisector of $\angle \mathrm{C}$.
The equation of the angle bisector is y = -x and the incenter 
$\mathrm{I}\equiv (-\mathrm{a},\mathrm{a})\text{, where }\mathrm{a}\text{ is positive. }$
$\text{ The equation of }\mathrm{AC}\text{ is }\mathrm{y}-0=-7(\mathrm{x}+6)\text{, i.e., }7\mathrm{x}+\mathrm{y}+42=0 \text{ and the equation of }\mathrm{AB}\text{ is }\mathrm{x}-\mathrm{y}+6=0\text{. }$
The lengths of the perpendiculars from I to AB and AC are equal. Therefore,
$\left|\frac{-7\mathrm{a}+\mathrm{a}+42}{\sqrt{50}}\right|=\left|\frac{-\mathrm{a}-\mathrm{a}+6}{\sqrt{2}}\right|$
$\mathrm{or} \mathrm{a}=\frac{9}{2} (\because \mathrm{a}>0)$
$\text{ Therefore, the center is }\left(-\frac{9}{2},\frac{9}{2}\right)\text{ and the radius is }\frac{3}{\sqrt{2}}$
Hence, the equation of the circle is
${\left(\mathrm{x}+\frac{9}{2}\right)}^2+{\left(\mathrm{y}-\frac{9}{2}\right)}^2=\frac{9}{2}$
$\text{ or }{\mathrm{x}}^2+{\mathrm{y}}^2+9\mathrm{x}-9\mathrm{y}+36=0$