Question

The locus of the midpoints of the chord of the circle $x^2+y^2=4$ which subtends a right angle at the origin is

• A. $x+y=2$
• B. $x^2+y^2=1$
• C. $x^2+y^2=2$
• D. $x+y=1$

Answer 1

The correct option is C

$x^2+y^2=2$

Explanation of the correct option.

Compute the locus:

Given: Chord subtends a right angle at the origin.

Let's plot the figure.

Since the perpendicular from the center bisects the chord.

In $∆OAC$,

$\frac{OC}{OA}=\sin \left(45°\right)$

$$\begin{gathered} \Rightarrow OC=2\times \frac{1}{\sqrt{2}} \\ \Rightarrow OC=\sqrt{2} \end{gathered}$$ $\left[\sin \left(45°\right)=\frac{1}{\sqrt{2}}\right]$

Using distance formula,

$$\begin{gathered} OC=\sqrt{{(h-0)}^2+{(k-0)}^2} \\ \Rightarrow \sqrt{2}=\sqrt{{(h-0)}^2+{(k-0)}^2} \\ \Rightarrow 2=h^2+k^2 \end{gathered}$$

Hence the locus is

$\Rightarrow x^2+y^2=2$

Option $\left(\mathrm{C}\right)$ is the correct option.