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

The circle with points P(2t₁,2/t₁) and Q(2t₂,2/t₂) as diameter is given by
$\left(\mathrm{x}-2{\mathrm{t}}_1\right)\left(\mathrm{x}-2{\mathrm{t}}_2\right)+\left(\mathrm{y}-\frac{2}{{\mathrm{t}}_1}\right)\left(\mathrm{y}-\frac{2}{{\mathrm{t}}_2}\right)=1$…………..(i)
Also, the slope of PQ is given by
$-\frac{1}{{\mathrm{t}}_1{\mathrm{t}}_2}=1\text{ or }{\mathrm{t}}_1{\mathrm{t}}_2=-1$
Hence, from (i), the circle is
$\left({\mathrm{x}}^2+{\mathrm{y}}^2-8\right)-\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)(\mathrm{x}-\mathrm{y})=0$
which is of the form $\mathrm{S}+\mathrm{\lambda L}=0$
Hence, circles pass through the points of intersection of the circle x²+y² -8 = 0and the line x= y.
The points of intersection are (2,2) and (-2,-2).