Question

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

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

Answer 1

The correct option is B

$x^2+y^2=2$

Explanation for the correct option

Step 1: Solve for radius of the given circle

Given, equation of circle is $x^2+y^2=4$ and it subtends an angle of $90°$ at the center.

We know that the equation of a circle centered at the origin is given as,
$x^2+y^2=r^2$
where $r$ is the radius of the circle

Comparing the given equation with the general form, we have,
$\begin{array}{cc}& r^2=4\\ \Rightarrow & r=2\end{array}$

Step 2: Solve for the required locus

Let the midpoint of the chord $AB$ be $C$.
The radius of the circle, $OB=2$

We have,
$\begin{array}{cc}& \sin \left(\frac{\mathrm{\pi }}{4}\right)=\frac{BC}{OB}\\ \Rightarrow & \frac{1}{\sqrt{2}}=\frac{BC}{2}\\ \Rightarrow & BC=\sqrt{2}\end{array}$

Let the coordinates of $C$ be $\left(x_1,y_1\right)$

Then the length of $OC=\sqrt{{x_1}^2+{y_1}^2}$ (Since the circle is centered at origin)

From Pythagoras theorem,

$\begin{array}{cc}& O{B}^2=O{C}^2+B{C}^2\\ \Rightarrow & 2^2={\left(\sqrt{{x_1}^2+{y_1}^2}\right)}^2+{\left(\sqrt{2}\right)}^2\\ \Rightarrow & {x_1}^2+{y_1}^2=2\end{array}$

Hence, option(B) i.e. $x^2+y^2=2$ is correct.