Circles are drawn on the chords of rectangular hyperbola xy = 4 parallel to line y = x as diameters. All such circles pass through two fixed points whose coordinates are

$\text{ Circles are drawn......... }$

$\text{ The circle with points }\left(2t_1,2/t_1\right)\text{ and }\left(2t_2,2/t_2\right)\text{ as diameter is }$

$\begin{array}{l}\left(x-2t_1\right)\left(x-2t_2\right)+\left(y-2/t_1\right)\left(y-2/t_2\right)=0\\ \text{ Also slope }-\frac{1}{t_1{t}_2}=1t_1{t}_2=-1\\ \text{ So equation of circle will become }\end{array}$

$\begin{array}{l}\left(x^2+y^2-8\right)-\left(t_1+t_2\right)(x-y)=0\\ \text{ which is of form }\mathrm{S}+\lambda \mathrm{L}=0\end{array}$

$\begin{array}{l}\text{ so two fixed points are intersection points of }(x-y)=0\\ \text{ and }{x}^2+y^2=8\\ (2,2)\mathrm{\&}(-2,-2)\end{array}$