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

The correct option is D

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

Explanation for correct answer:

Step-1 Point of Intersections of ellipse and line:

Given: $\frac{x^2}{4}+y^2=1$ intersects $x=2y$ at $P$ and $Q$.

Put $x=2y$ in the equation $\frac{x^2}{4}+y^2=1$

$\begin{array}{cc}\Rightarrow & \frac{{\left(2y\right)}^2}{4}+y^2=1\\ \Rightarrow & \frac{4y^2}{4}+y^2=1\\ \Rightarrow & y^2+y^2=1\\ \Rightarrow & 2y^2=1\\ \Rightarrow & y^2=\frac{1}{2}\\ \Rightarrow & y=\pm \frac{1}{\sqrt{2}}\end{array}$

$$\begin{gathered} \Rightarrow x=2y \\ \Rightarrow x=2\times \pm \frac{1}{\sqrt{2}} \\ \Rightarrow x=\pm \sqrt{2} \end{gathered}$$

So, points $P=\left(\sqrt{2},\frac{1}{\sqrt{2}}\right),Q=\left(-\sqrt{2},-\frac{1}{\sqrt{2}}\right)$

Step-2 Equation of circle:

Equation of circle having a diameter at points $(a,b),(c,d)$ is $(x-a)(x-c)+(y-b)(y-d)=0$

Equation of circle having a diameter at points $P=\left(\sqrt{2},\frac{1}{\sqrt{2}}\right),Q=\left(-\sqrt{2},-\frac{1}{\sqrt{2}}\right)$ is

$$\begin{gathered} \Rightarrow (x-\sqrt{2})(x+\sqrt{2})+(y-\frac{1}{\sqrt{2}})(y+\frac{1}{\sqrt{2}})=0 \\ \Rightarrow x^2-2+y^2-\frac{1}{2}=0 \\ \Rightarrow x^2+y^2-\frac{5}{2}=0 \\ \Rightarrow x^2+y^2=\frac{5}{2} \end{gathered}$$

Hence, option D is correct.