Two straight rods of lengths $2\mathrm{a}$ and $2\mathrm{b}$ move along the coordinate axes in such a way that their extremities are always concyclic. Then the locus of the centre of such circles is

$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0 \mathrm{be} \mathrm{circle} \\ \mathrm{centre} \left(\mathrm{c}\right)=(-\mathrm{g}, -\mathrm{f}) \\ \mathrm{Given} \mathrm{x}-\mathrm{intercert}=2\mathrm{a} \\ \Rightarrow 2\sqrt{{\mathrm{g}}^2-\mathrm{c}}=2\mathrm{a} \\ \Rightarrow {\mathrm{g}}^2-\mathrm{c}={\mathrm{a}}^2 \\ \Rightarrow {\mathrm{g}}^2-{\mathrm{a}}^2=\mathrm{c} - \left(1\right) \\ \mathrm{Given} \mathrm{y}-\mathrm{intercert}=2\mathrm{b} \\ \Rightarrow 2\sqrt{{\mathrm{f}}^2-\mathrm{c}}=2\mathrm{b} \\ \Rightarrow {\mathrm{f}}^2-\mathrm{c}={\mathrm{b}}^2 \\ \Rightarrow {\mathrm{f}}^2-{\mathrm{b}}^2=\mathrm{c} - \left(2\right) \\ \mathrm{from} \left(1\right) \& \left(2\right) {\mathrm{g}}^2-{\mathrm{a}}^2={\mathrm{f}}^2-{\mathrm{b}}^2 \\ \Rightarrow {\mathrm{g}}^2-{\mathrm{f}}^2={\mathrm{a}}^2-{\mathrm{b}}^2 \\ \therefore \mathrm{locus} \mathrm{of} \mathrm{centre} (-\mathrm{g}, -\mathrm{f}) \mathrm{is} {\mathrm{x}}^2-{\mathrm{y}}^2={\mathrm{a}}^2-{\mathrm{b}}^2 \end{gathered}$$