Question

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

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

Solving $x=2y$ ..... (i)
and

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

Put $x=2y$ in Eq. (ii), we get
$\frac{(2y{)}^2}{4}+y^2=1\Rightarrow \frac{4y^2}{4}+y^2=1$
$\Rightarrow 2y^2=1\Rightarrow y=\pm \frac{1}{\sqrt{2}}$

$\therefore$ From Eq. (i), $x=\pm \sqrt{2}$

$\therefore P(\sqrt{2},\frac{1}{\sqrt{2}})$ and $Q(-\sqrt{2},-\frac{1}{\sqrt{2}})$

$\therefore$ Equation of circle with $PQ$ as diameter is

$(x-\sqrt{2})(x+\sqrt{2})+(y-\frac{1}{\sqrt{2}})(y+\frac{1}{\sqrt{2}})$

$\Rightarrow x^2-2+y^2-\frac{1}{2}=0\Rightarrow x^2+y^2=\frac{5}{2}$