Question

The locus of the midpoint of a chord of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=4$ which subtends a right angle at the origin is

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

Answer 1

The correct option is C

${\mathrm{x}}^2+{\mathrm{y}}^2=2$

Explanation of the correct answer.

Compute the locus:

Given: ${\mathrm{x}}^2+{\mathrm{y}}^2=4$

From the equation, radius of the circle is $2$ and the center is $\left(0,0\right)$.

Let $\mathrm{P}(\mathrm{h},\mathrm{k})$ is the midpoint of the chord.

Distance of $\mathrm{OP}=\sqrt{{\left(\mathrm{h}-0\right)}^2+{\left(\mathrm{k}-0\right)}^2}$ $...\left[Distance=\sqrt{{\left(x_2-x_1\right)}^2+{(y_2-y_1)}^2}\right]$

Since chord making right angle at origin $\mathrm{O}\left(0,0\right)$ and $\mathrm{P}(\mathrm{h},\mathrm{k})$ is midpoint, $\mathrm{OP}$ will bisect the right angle.

$$\begin{gathered} \mathrm{OP}=\mathrm{r}\cos \left(45°\right) \\ \Rightarrow \sqrt{{\left(\mathrm{h}-0\right)}^2+{\left(\mathrm{k}-0\right)}^2}=2\times \left(\frac{1}{\sqrt{2}}\right) \\ \Rightarrow \sqrt{{\left(\mathrm{h}-0\right)}^2+{\left(\mathrm{k}-0\right)}^2}=\sqrt{2} \\ \Rightarrow {\mathrm{h}}^2+{\mathrm{k}}^2=2 \end{gathered}$$

Therefore, the locus of $\mathrm{P}(\mathrm{h},\mathrm{k})$ is ${\mathrm{x}}^2+{\mathrm{y}}^2=2$.

Hence option $\mathrm{C}$ is the correct option.