Two rods of lengths $aandb$ slide along the $x-axis$ and $y-axis$  respectively in such a manner that their ends are concyclic. The locus of the centre of the circle passing through the end points is

Let $x^2-y^2=a^2-b^2$ be the centre of the circle passing through the end points of the rod AB and PQ of lengths $aandb$  respectively, CL and CM be perpendiculars from C on AB and PQ respectively.

then $AL=\left(1/2\right)AB=a/2$

$PM=\left(1/2\right)PQ=b/2$

and CA =CP(radii of the same circle)

$\Rightarrow k^2+\frac{a^2}{4}=h^2+\frac{b^2}{4}$

$\Rightarrow 4\left(h^2-k^2\right)=a^2-b^2$

so that locus of $\left(h,k\right)$ is $4\left(x^2-y^2\right)=a^2-b^2$