Question

A tangent to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ at any point $P$ meets the line $x=0$ at a point $Q.$ Let $R$ be the image of $Q$ in the line $y=x$, then the circle whose extremities of a diameter are $Q$ and $R$ passes through a fixed point. The fixed point is

• A. $(3,0)$
• B. $(5,0)$
• C. $(0,0)$
• D. $(4,0)$

Answer 1

The correct option is C $(0,0)$

Equation of the tangent to the ellipse at $P(5\cos \theta ,4\sin \theta )$ is
$\frac{x\cos \theta }{5}+\frac{y\sin \theta }{4}=1.$

It meets the line $x=0$ at $Q(0,4\text{cosec}\theta )$
Image of $Q$ in the line $y=x$ is $R(4\text{cosec}\theta ,0)$

Equation of the circle whose extremities of diameter are $Q$ and $R,$ is given by :
$x(x-4\text{cosec}\theta )+(y-4\text{cosec}\theta )y=0$
$\Rightarrow x^2+y^2-4(x+y)\text{cosec}\theta =0$

It represents a family of circle, which always passes through the point of intersection of circle $x^2+y^2=0$ and line $x+y=0,$ which is given by $(0,0).$