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

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$x^{2} + y^{2} - 9 x - 9 y + 36 = 0$

$x^{2} + y^{2} + 18 x - 18 y + 36 = 0$

$x^{2} + y^{2} + 9 x - 9 y + 36 = 0$

$x^{2} + y^{2} + 9 x + 9 y - 36 = 0$

answer is B.

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Detailed Solution

The Triangle is evidently isosceles and , therefore ,the median through C is the angle bisector of $∠$ C
The equation of the angle bisector is y = -X and the incenter
I $\equiv$ (-a, a), where a is Positive'

The equation of AC is y-0 =-7(x+6), 7x+3y+42=0
and the equation of AB is x-y+6=0
The perpendiculars From I to AB and AC are equal. therefore,
$|\frac{- 7 a + a + 42}{\sqrt{50}}| = |\frac{- a - a + 6}{\sqrt{2}}|$
$a = \frac{9}{2}$ $(∵ a > 0)$
Therefore the centre is $(- 9 / 2, 9 / 2)$ and the radius is $3 \sqrt{2}$
Hence, the equation of the circle is
${(x + \frac{9}{2})}^{2} + {(y - \frac{9}{2})}^{2} = \frac{9}{2}$
or $x^{2} + y^{2} + 9 x - 9 y + 36 = 0$