Question

The line $x=2 y$ intersects the ellipse $\frac{x^{2}}{4}+y^{2}=1$ at the points $P$ and $Q$. The equation of the circle with $P Q$ as diameter is

• A. $x^{2}+y^{2}=\frac{1}{2}$
• B. $x^{2}+y^{2}=1$
• C. $x^{2}+y^{2}=2$
• D. $x^{2}+y^{2}=\frac{5}{2}$

Answer 1

Answer: (D)

$x^{2}+y^{2}=\frac{5}{2}$

Answer 2

Solving $x=2 y$ ...(i)
and $\frac{x^{2}}{4}+y^{2}=1$ ...(ii)

Put $x=2 y$ in E (ii), we get
$\frac{(2 y)^{2}}{4}+y^{2}=1$
$\Rightarrow \frac{4 y^{2}}{4}+y^{2}=1$
$\Rightarrow 2 y^{2}=1$
$\Rightarrow y=\pm \frac{1}{\sqrt{2}}$
$\therefore$ From E (i), $x=\pm \sqrt{2}$
$\therefore P\left(\sqrt{2}, \frac{1}{\sqrt{2}}\right)$
and $Q\left(-\sqrt{2},-\frac{1}{\sqrt{2}}\right)$
$\therefore$ Equation of circle with $P Q$ as diameter is
$(x-\sqrt{2})(x+\sqrt{2})+\left(y-\frac{1}{\sqrt{2}}\right)\left(y+\frac{1}{\sqrt{2}}\right)$
$\Rightarrow x^{2}-2+y^{2}-\frac{1}{2}=0$
$\Rightarrow x^{2}+y^{2}=\frac{5}{2}$