Circles are drawn on 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
Step-by-Step Solution
Key Concept: For a circle with diameter PQ where P and Q lie on xy=4 with chord slope -1, the constraint t₁t₂=-1 (from slope condition) makes the circle equation x²+y²-8-2(t₁+t₂)(x-y)=0 independent of the parameter (t₁+t₂) at the fixed points, which satisfy both the circle and hyperbola simultaneously.
A circle with diameter $PQ$ where $P(2t_1, 2/t_1)$ and $Q(2t_2, 2/t_2)$ lie on the rectangular hyperbola $xy=2$ is formed. Using the diameter condition and the fact that slope of $PQ$ equals $-1/t_1t_2$, we get $t_1t_2 = -1$. The circle equation becomes $x^2 + y^2 - 8 - 2(t_1 + t_2)(x - y) = 0$, which can be written as $S + \lambda L = 0$. Finding intersection with line $x = y$ yields points $(2, 2)$ and $(-2, -2)$.
Correct Answer: 1,4