Question

The circle $x^2+y^2-4x-4y+4=0$ is inscribed in a traingle which has two sides along the coordianates axes. 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. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is A $1$

The given circle is

$⟹$ centre $=(2,2)$ and radius $=2$

Let $OAB$ be the triangle in which the circle is inscribed.

As $△OAB$ is right angled, the circumcentre is mid-point of $AB$.

Let $P\equiv (x_1,y_1)$ be the circumcentre.

$⟹A\equiv (2x_1,0)$ and $B\equiv (0,2y_1)$

$⟹$ Equation of $AB$ is $\frac{x}{2x_1}+\frac{y}{2y_1}=1$

As $△OAB$ touches the circle, distance of centre $C$ from $AB$ is the radius.

$⟹\frac{|\frac{2}{2x_1}+\frac{2}{2y_1}-1|}{\sqrt{\frac{1}{4{x_1}^2}+\frac{1}{4{y_1}^2}}}=2$ ....... $\left(i\right)$

As the centre $(2,2)$ lies on the origin side of the line $\frac{x}{2x_2}+\frac{y}{2y_1}-1=0$, the equation $\frac{2}{2x_2}+\frac{2}{2y_1}-1$ has the same sign as the constant term $(-1)$ in the equation

$⟹\frac{2}{2x_2}+\frac{2}{2y_1}-1$ is negative

Equation $\left(i\right)$ becomes

$⟹-(\frac{2}{2x_1}+\frac{2}{2y_1}-1)=2\sqrt{\frac{1}{4{x_1}^2}+\frac{1}{4{y_1}^2}}$

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

$⟹x_1+y_1-x_1{y}_1+\sqrt{{x_1}^2+{y_1}^2}=0$ is the locus

$⟹m=1$