Question

$(6,0),(0,6)$ and $(7,7)$ are the vertices of a triangle. The circle inscribed in the triangle has 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. $x^2+y^2-9x-9y-36=0$

Answer 1

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

Let, $△ABC$ has the vertices $A(6,0),B(0,6),C(7,7)$ and $AB,BC,CA$ are the sides with distances $c,a,b$ respectively.

$\therefore$ Using distance formula, $a=\left|BC\right|=\sqrt{{(7-0)}^2+{(7-6)}^2}=5\sqrt{2}b=\left|CA\right|=\sqrt{{(7-6)}^2+{(7-0)}^2}=5\sqrt{2}c=\left|AB\right|=\sqrt{{(6-0)}^2+{(0-6)}^2}=6\sqrt{2}$

Now, assuming the In-centre of the circle as $O$ we get,

$O=(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})$

$=(\frac{5\sqrt{2}\left(6\right)+5\sqrt{2}\left(0\right)+6\sqrt{2}\left(7\right)}{5\sqrt{2}+5\sqrt{2}+6\sqrt{2}},\frac{5\sqrt{2}\left(0\right)+5\sqrt{2}\left(6\right)+6\sqrt{2}\left(7\right)}{5\sqrt{2}+5\sqrt{2}+6\sqrt{2}})$

$\Rightarrow O=(\frac{9}{2},\frac{9}{2})$

and,

The inradius r is equal to the perpendicular distance from incenter to any of the sides

Equation of the line $AB$

$y-0=\frac{6-0}{0-6}(x-6)$ $[two-pointform]$

$y=\frac{6}{-6}(x-6)$

$y=-(x-6)$

$x+y-6=0$

So the inradius is

$r=\frac{|\frac{9}{2}+\frac{9}{2}-6|}{\sqrt{1+1}}$

-

$r=\frac{|9-6|}{\sqrt{2}}$

$r=\frac{|3|}{\sqrt{2}}$

$\therefore r=\frac{|3|}{\sqrt{2}}$

The equation of the incircle is

$(x-\frac{9}{2}{)}^2+(y-\frac{9}{2}{)}^2={\left(\frac{|3|}{\sqrt{2}}\right)}^2$

$x^2+\frac{81}{4}-9x+y^2+\frac{81}{4}-9y=\frac{9}{2}$

$x^2+-9x+y^2-9y+\frac{81}{4}+\frac{81}{4}-\frac{9}{2}=0$

$x^2+y^2-9x-9y+\frac{81}{2}-\frac{9}{2}=0$

$x^2+y^2-9x-9y+\frac{81-9}{2}=0$

$x^2+y^2-9x-9y+\frac{72}{2}=0$

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