If the circle ${\mathrm{x}}^2+{\mathrm{y}}^2= {\mathrm{r}}^2$ intersects the hyperbola $\mathrm{xy}={\mathrm{c}}^2$ in four points $({\mathrm{x}}_{\mathrm{i}}, {\mathrm{y}}_{\mathrm{i}})$ for i = 1,2,3 and 4 then ${\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3+{\mathrm{y}}_4=$

$$\begin{gathered} \mathrm{Given} \mathrm{circle} {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2 — ① \\ and hyperbola \mathrm{xy}={\mathrm{c}}^2 \\ \Rightarrow \mathrm{x}=\frac{{\mathrm{c}}^2}{\mathrm{y}} — ② \\ \mathrm{Now} \mathrm{P}.\mathrm{I} \mathrm{of} ① \& ② \mathrm{is} \\ {\left(\frac{{\mathrm{c}}^2}{\mathrm{y}}\right)}^2+{\mathrm{y}}^2={\mathrm{r}}^2 \\ \Rightarrow \frac{{\mathrm{c}}^4}{{\mathrm{y}}^2}+{\mathrm{y}}^2={\mathrm{r}}^2 \\ \Rightarrow {\mathrm{y}}^4-{\mathrm{r}}^2{\mathrm{y}}^2+{\mathrm{c}}^4=0 \\ \mathrm{It} \mathrm{has} \mathrm{four} \mathrm{roots} \mathrm{let} \mathrm{it} \mathrm{be} {\mathrm{y}}_1, {\mathrm{y}}_2, {\mathrm{y}}_3, {\mathrm{y}}_4 \\ \mathrm{Now} {\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3+{\mathrm{y}}_4=0 \end{gathered}$$