Explanation

(c)
Let $C(h,\ k)$ be the centre of the circle passing through the end points of the rod $AB$ and $PQ$ of lengths $a$ and $b$ respectively. $CL$ and $CM$ be perpendicular drawn from $C$ on $AB$ and $PQ$ respectively. Then, $CA = CP$ (radii of the same circle)
$\Rightarrow k + \frac{a^{2}}{4} = h^{2}\mathbf{+}\frac{b^{2}}{4}\mathbf{\ \ \ \ }\left( \because AL = \frac{a}{2}\ and\ MP = \frac{b}{2} \right)$
$\Rightarrow 4\left( h^{2} - k^{2} \right) = a^{2} - b^{2}$
[image:11_image69.emf]
Hence, locus of $(h,\ k)$ is $4\left( x^{2} - y^{2} \right) = a^{2} - b^{2}$