If a variable circle ${\mathrm{x}}^2 +{\mathrm{y}}^2 –2\mathrm{ax}+4\mathrm{ay}=0$ intersect the hyperbola xy=4 at the points $({\mathrm{x}}_{\mathrm{i}}, {\mathrm{y}}_{\mathrm{i}} ) \mathrm{i} = 1, 2, 3, 4$ then the locus of the point $\left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4}{4},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3+{\mathrm{y}}_4}{4}\right)$ is

$$\begin{gathered} \mathrm{Given} \mathrm{circle} \mathrm{and} \mathrm{hyperbola} \mathrm{are} \\ {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{ax}+4\mathrm{ay}=0 — ① \\ \mathrm{xy}=4 \\ \Rightarrow \mathrm{y}=\frac{4}{\mathrm{x}} \\ \mathrm{sub} \mathrm{y} in ① \\ {\mathrm{x}}^2+\frac{16}{{\mathrm{x}}^2}-2\mathrm{ax}+4\mathrm{a}\left(\frac{4}{\mathrm{x}}\right)=0 \\ \Rightarrow {\mathrm{x}}^4-2{\mathrm{ax}}^3+16\mathrm{ax}+16=0 \\ \mathrm{Now} {\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4=2\mathrm{a} \\ \mathrm{xy}=4\Rightarrow \mathrm{x}=\frac{4}{y} \\ \mathrm{sub} \mathrm{x} in ① \\ \frac{16}{{\mathrm{y}}^2}+y^2-\frac{8a}{y}+4ay=0 \\ \Rightarrow y^4+4a{y}^3-8ay+16=0 \\ \mathrm{Now} {\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3+{\mathrm{y}}_4=-4\mathrm{a} \end{gathered}$$

$$\begin{gathered} \mathrm{let}(\mathrm{h}, \mathrm{k})=\left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4}{4}, \frac{{\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3+{\mathrm{y}}_4}{4}\right) \\ \Rightarrow {\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4=4\mathrm{h}\Rightarrow 4\mathrm{h}=2\mathrm{a} — ② \\ \Rightarrow {\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3+{\mathrm{y}}_4=4\mathrm{k}\Rightarrow 4\mathrm{k}=-4\mathrm{a} \Rightarrow k=-a — ③ \\ \mathrm{from} ② \& ③ 4\mathrm{h}=-2\mathrm{k} \\ \therefore \mathrm{locus} \mathrm{of} (\mathrm{h}, \mathrm{k}) \mathrm{is} 2\mathrm{x}+\mathrm{y}=0 \end{gathered}$$