Two rods of lengths ‘a’ and ‘b’ slide along coordinate axes, such that their ends are concyclic. Locus of centre of circle is

$Let the circle be x^2+y^2+2gx+2fy+c=0$

Length of intercept on x-axis is $=2\sqrt{{\mathrm{g}}^2-\mathrm{c}}$

and on y-axis  is $=2\sqrt{{\mathrm{f}}^2-\mathrm{c}}$

$$\begin{gathered} \mathrm{Given} \mathrm{that} \\ \mathrm{a}=2\sqrt{{\mathrm{g}}^2-\mathrm{c}} \\ \mathrm{b}=2\sqrt{{\mathrm{f}}^2-\mathrm{c}} \\ Now {\mathrm{a}}^2-{\mathrm{b}}^2=4({\mathrm{g}}^2-\mathrm{c})-4({\mathrm{f}}^2-\mathrm{c}) \\ \mathrm{Now} \mathrm{locus} \mathrm{of} \mathrm{centre} (-\mathrm{g},-\mathrm{f}) \mathrm{is} \\ \Rightarrow {\mathrm{a}}^2-{\mathrm{b}}^2=4({\mathrm{x}}^2-\mathrm{c})-4({\mathrm{y}}^2-\mathrm{c}) \\ =4({\mathrm{x}}^2-{\mathrm{y}}^2) \\ \Rightarrow {\mathrm{a}}^2-{\mathrm{b}}^2=4({\mathrm{x}}^2-{\mathrm{y}}^2) \end{gathered}$$