Question

The circle $x^{2}+y^{2}-4 x-4 y+4=0$ 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 $x+y-x y +k \sqrt{x^{2}+y^{2}}=0$, then the value of $k$ is equal to

• A. 2
• B. 1
• C. -2
• D. 3

Answer 1

Answer: (B)

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

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