Question

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

• A. $x^2+y^2-9x-9y+36=0$
• B. $x^2+y^2+9x-9y+36=0$
• C. $x^2+y^2+9x+9y-36=0$
• D. $x^2+y^2+18x-18y+36=0$

Answer 1

Answer: (B)

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

The triangle is evidently isosceles and therefore the median through C is the angle bisector of $\angle C$

The equation of the angle bisector is $y=-x$ and in centre $I=(-a,a)$ where a is positive.

Equation of AC is $y-0=-7(x+6)$ or $7x+y+42=0$ and equation of AB is $x-y+6=0$.

The length of the perpendicular from I to BC and AC are equal.

$\therefore \left|\frac{-7+a+42}{\sqrt{50}}\right|=\left|\frac{-a-a+6}{\sqrt{2}}\right|$

$\therefore a=\frac{9}{2}(\because a>0)$

Centre is $(-\frac{9}{2},\frac{9}{2})$ and radius $=\frac{3}{\sqrt{2}}$

$\therefore$ The equation of the circle is ${(x+\frac{9}{2})}^2+{(y-\frac{9}{2})}^2=\frac{9}{2}$

$\therefore x^2+y^2+9x-9y+36=0$