Question

The locus of the mid-points 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 $x^2+y^2=2$

As, we have to find the locus of mid-point of chord and we know perpendicular drawn from centre to the chord, it bisects the chord.

Given, $\angle OAB={90}^{\circ }$$

Also, $OC$ bisects the $\angle AOB$

$\therefore \angle COA=\angle OAC={45}^{\circ }$

In $△OAC,\frac{OC}{OA}=\sin {45}^{\circ }\Rightarrow OC=\frac{2}{\sqrt{2}}=\sqrt{2}$

Therefore, $h^2+k^2=O{C}^2$

$\Rightarrow h^2+k^2=(\sqrt{2}{)}^2$

Thus $x^2+y^2=2$ is required equation of locus of mid-point of chord subtending right angle at centre.