Question

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

• A. $1$
• B. $-1$
• C. $2$
• D. None of these

Answer 1

Answer: (A), (B)

$-1$
$1$

Let $OAB$ be the triangle in which the given circle of radius $2$ with centre $C(2,2)$ is inscribed and let the equation of $AB$ be

$\frac{x}{a}+\frac{y}{b}=1$ ...(1)

$\therefore OA=a,OB=b$

$\Rightarrow$ the centre of the circumscribed circle of $△OAB$ is $(\frac{a}{2},\frac{b}{2}).$

Let $\alpha =\frac{a}{2}$ and $\beta =\frac{b}{2}$

Now, the length of the $\perp$ from $C(2,2)$ on (1) must be equal to the radius $2.$

$\therefore \pm 2=\frac{|\frac{2}{a}+\frac{2}{b}-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$

Eliminating $a$ and $b$ from (2) and (3), we get

$\therefore \pm 2=\frac{|\frac{1}{\alpha }+\frac{1}{\beta }-1|}{\sqrt{\frac{1}{{4\alpha }^2}+\frac{1}{{4\beta }^2}}}\Rightarrow \pm 1=\frac{\alpha +\beta -\alpha \beta }{\sqrt{{\alpha }^2+{\beta }^2}}$

$\Rightarrow \alpha +\beta -\alpha \beta \pm \sqrt{{\alpha }^2+{\beta }^2}=0$

$\therefore$ Locus of $(\alpha ,\beta )$ is $x+y-xy\pm \sqrt{x^2+y^2}=0.$

$\therefore k=\pm 1.$