Question

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

• A. $(2,2)$
• B. $(2,-2)$
• C. $(-2,2)$
• D. $(-2,-2)$

Answer 1

Answer: (A), (D)

$(-2,-2)$

Circle with points$P(2t_1,\frac{2}{t_1})$ and $Q(2t_2,\frac{2}{t_2})$ as diameter is given by

$(x-2t_1)(x-2t_2)+(y-\frac{2}{t_1})(y-\frac{2}{t_2})=0\cdots \left(1\right)$

Also, slope of $PQ$ is given by

$-\frac{1}{t_1{t}_2}=1\Rightarrow t_1{t}_2=-1$

Hence, from $\left(1\right)$ ,circle is

$(x^2+y^2-8)-2(t_1+t_2)(x-y)=0$

which is of the form $S+\lambda L=0$

Hence, circles pass through the points of intersection of the circle $x^2+y^2-8=0$ and the line $x=y$

The points of intersection are $(2,2)$ and $(-2,-2)$