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

The correct option is B $x^2+y^2$ + 9x – 9y + 36 = 0

The triangle is isosceles, therefore the median through C is the angle bisector of $\angle C$. Equation of angle bisector is y = –x, thus, incenter I $\equiv$ (–a, a); a > 0
Equation of AC $\equiv$ 7x + y + 42 = 0 and AB $\equiv$ x – y + 6 = 0
Length of the perpendicular from I to AB and AC are equal.
$\therefore \left|\frac{-7a+a+42}{\sqrt{50}}\right|=\left|\frac{-a-a+6}{\sqrt{2}}\right|\Rightarrow a=\frac{9}{2}\therefore center\equiv (-\frac{9}{2},\frac{9}{2})andradius=\frac{3}{\sqrt{2}}$
$\therefore$ The equation of the circle is
${(x+\frac{9}{2})}^2+{(y-\frac{9}{2})}^2=\frac{9}{2}\Rightarrow x^2+y^2+9x-9y+36=0$