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 whose coordinates are

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

Answer 1

Answer: (C)

$(0,0)$

$\frac{x^2}{25}+\frac{y^2}{16}=1$

$P(5\cos \theta ,4\sin \theta )$

Tangent $\frac{x}{5}\cos \theta +\frac{y}{4}\sin \theta =1$

$Q(\theta ,\frac{4}{\sin \theta })$

$R(\frac{4}{\sin \theta },\theta )$

Cicle with $QR$ as diameter

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

$\Rightarrow x(x-\frac{4}{\sin \theta })+y(y-\frac{4}{\sin \theta })=0$

passes through $(0,0)$