Question

Let $L_1:x+y=1$ be a line and two perpendicular lines (which are not parallel or perpendicular to $L_1$) passes through $(2,1)$. If $C$ is incircle of the triangle formed by the three lines, then the locus of incentre of $C,$ is

• A. $xy+x-4=0$
• B. $xy+y-2=0$
• C. $xy+x-2=0$
• D. $y^2+x^2-xy-2=0$

Answer 1

The correct option is C $xy+x-2=0$

Let $C(h,k)$ be the centre and $r$ be the radius of the circle.


As $\angle APB={90}^{\circ }\Rightarrow PACB$ is a square.
$\Rightarrow CP=\sqrt{2}r$
$\Rightarrow (h-2{)}^2+(k-1{)}^2=2r^2$ $\left(1\right)$
Now, $x+y=1$ is a tangent to the circle
$\Rightarrow \frac{|h+k-1|}{\sqrt{2}}=r$
$\Rightarrow (h+k-1{)}^2=2r^2$ $\left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$,
$h^2-4h+4+k^2-2k+1=h^2+k^2+1+2hk-2k-2h$
$\Rightarrow 2hk=-2h+4$
$\Rightarrow hk+h-2=0$

Hence the locus of the incentre will be
$\Rightarrow xy+x-2=0$