If the circle $${x^2} + {y^2} = {a^2}$$   intersects the hyperbola $$xy = {c^2}$$   in four points $$P\left( {{x_1},\,{y_1}} \right),\,Q\left( {{x_2},\,{y_2}} \right),\,R\left( {{x_3},\,{y_3}} \right),\,S\left( {{x_4},\,{y_4}} \right){\text{ then :}}$$

Solution :
$$\eqalign{ & \left( {{x_{\text{i}}},\,{y_{\text{i}}}} \right),\,{\text{i}} = 1,\,2,\,3,\,4{\text{ lies on}} \cr & xy = {c^2} \Rightarrow {y_{\text{i}}} = \frac{{{c^2}}}{{{x_{\text{i}}}}} \cr & {\text{Now the point }}\left( {{x_{\text{i}}},\,{y_{\text{i}}}} \right){\text{ lies on}} \cr & {x^2} + {y^2} = {a^2} \cr & \Rightarrow x_{\text{i}}^2 + \frac{{{c^4}}}{{{x_{\text{i}}}}} = {a^2} \cr & \Rightarrow x_{\text{i}}^4 - {a^2}x_{\text{i}}^2 + {c^4} = 0 \cr & {\text{Its roots are }}{x_1},\,{x_2},\,{x_3},\,{x_4} \cr & \therefore \,{x_1} + {x_2} + {x_3} + {x_4} = 0 \cr & \Rightarrow {x_1}{x_2} + {x_1}{x_3} + {x_1}{x_4} + {x_2}{x_3} + {x_2}{x_4} + {x_3}{x_4} = {a^2} \cr & \Rightarrow {x_1}{x_2}{x_3} + {x_1}{x_2}{x_4} + {x_1}{x_3}{x_4} + {x_2}{x_3}{x_4} = 0 \cr & \Rightarrow {x_1}{x_2}{x_3}{x_4} = {c^4}{\text{ Clearly }}\left( {\bf{C}} \right){\text{ is not correct}} \cr & {\text{Now, }}{y_1} + {y_2} + {y_3} + {y_4} = \frac{{{c^2}}}{{{x_1}}}.\frac{{{c^2}}}{{{x_2}}}.\frac{{{c^2}}}{{{x_3}}}.\frac{{{c^2}}}{{{x_4}}} = {c^4} \cr & {\text{and }}{y_1} + {y_2} + {y_3} + {y_4} = \frac{{{c^2}\left( {\sum {{x_1}{x_2}{x_3}} } \right)}}{{{x_1}{x_2}{x_3}{x_4}}} = 0 \cr} $$