The locus of the mid points of the chords of the circle $x^{2}+y^{2}=4$ which subtend a right angle at the origin is

• A

  $x^{2}+y^{2}=1$
• Correct answer
  B

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

  $x+y=1$
• D

  $x+y=2$

Solution

Let mid point of the chord $\mathrm{AB}$ is $\mathrm{C}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$. In $\Delta \mathrm{COB}, \sin \frac{\pi}{4}=\frac{\mathrm{BC}}{\mathrm{OB}}$

$$ \begin{array}{ll} \Rightarrow & \frac{1}{\sqrt{2}}=\frac{\mathrm{BC}}{2} \\ \Rightarrow & \mathrm{BC}=\sqrt{2} \end{array} $$ Using Pythagoras theorem $\mathrm{OB}^{2}=\mathrm{OC}^{2}+\mathrm{CB}^{2}$ $\Rightarrow \quad(2)^{2}=\mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}+(\sqrt{2})^{2}$ $\Rightarrow \quad \mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}=2$ Hence, locus of mid point of chord is $$ x^{2}+y^{2}=2 $$

Hence, the correct answer is option: 'B'.