Explanation

(b)
Let the equation of $AB$ be $\frac{x}{a} + \frac{x}{b} = 1$
Since, the line $AB$ touches the circle
$x^{2} + y^{2} - 4x - 4y + 4 = 0$
$\therefore\ \ \ \frac{\left| \frac{2}{a} + \frac{2}{b} - 1 \right|}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} = 2$
[image:11_image56.emf]
[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, $\mathrm{\Delta}OAB$ is a right angled triangle. So, its circumcentre is the mid point of $AB$
$\therefore\ \ h = \frac{a}{2}and\ \ k = \frac{b}{2}\ \Rightarrow \ \ \ a = 2h\ \ \ and\ \ \ b = 2k$ …(ii)
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
$x + y - xy + \sqrt{x^{2} + y^{2}} = 0\ \ \ \therefore k = 1$