Question

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

• A. 2
• B. 1
• C. -2
• D. 3

Answer 1

Answer: (B)

Let the equation of $AB$ be $\frac{xb}{a}+\frac{y}{b}=1.$
Since, the line $AB$ touches the circle
$x^2+y^2-4x-4y+4=0$
$\therefore \frac{|\frac{2}{a}+\frac{2}{b}=1|}{\sqrt{\frac{t}{a^2}+\frac{1}{b^2}}}=2$
[Since, $O(0,0)$ and $C(2,2)$ lie on the same side of $AB$, therefore $\frac{2}{a}+\frac{2}{b}-1<0]$

$\Rightarrow \frac{-(2b+2a-ab)}{\sqrt{a^2+b^2}}=2$
$\Rightarrow 2a-2b-ab+2\sqrt{a^2+b^2}=0$ ... (i)
Since, $\Delta OAB$ is a right angled triangle.
So, its circumcentre is the midpoint of $AB$.
$\therefore h=\frac{a}{2}$ and $k=\frac{b}{2}$ ... (ii)
$\Rightarrow a=2h$ and $b=2k$
From Eqs. (i) and (ii), we get
$4h+4k-4hk+2\sqrt{4h^2+4k^2}=0$
$\Rightarrow h+k-hk+\sqrt{h^2+k^2}=0$
So, the locus of $P(h,k)$ is
$x+y-xy+\sqrt{x^2+y^2}=0$
But the locus of the circumcentre is given to be
$x+y-xy+k\sqrt{x^2+y^2}=0$
$\therefore k=1$