Question

The circle $x^2+y^2-4x-4y+4=0$ is insrcibed in a triangle which has two sides along the coordinate axes. If the locus of the circumcentre of the triangle is $x+y-xy+m\sqrt{x^2+y^2}=0$, then $m$ is equal to

• A. $1$
• B. $3$
• C. $5$
• D. None

Answer 1

The correct option is A $1$

If $(x_1,y_1)$ is the circumcentre of the $△ABO,$

then $(x_1,y_1)$ is mid-point of the hypotenuse $AB.$

Therefore, $A$ is $(2x_1,0)$ and $B$ is $(0,2y_1).$

Also, $x_1+y_1{-x}_1{y}_1+m\sqrt{x_1^2+y_1^2}=0$ ...(1)

The inradius $=\sqrt{2^2+2^2-4}=2$

i.e., $\frac{Areaof△ABC}{\left(\frac{OA+OB+AB}{2}\right)}=2$ $[$Using $r=\frac{△}{s}]$

$\Rightarrow \frac{\frac{1}{2}\left(2x_1\right)\left(2y_1\right)}{x_1+y_1+\sqrt{x_1^2+y_1^2}}=2$

$\Rightarrow x_1+y_1-x_1{y}_1+\sqrt{x_1^2+y_1^2}=0$ ...(2)

From Eq. (1) and (2), we get $m=1.$