question_answer

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)        \[x+y=1\]

Correct Answer: C

Solution :

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,   \[\angle OAB={{90}^{o}}\] Also, \[OC\]bisects the\[\angle AOB\] \[\therefore \]  \[\angle COA=\angle OAC={{45}^{o}}\] In\[\Delta OAC,\,\,\frac{OC}{OA}=\sin {{45}^{o}}\]    \[\Rightarrow \]    \[OC=\frac{2}{\sqrt{2}}=\sqrt{2}\] \[\therefore \]  \[{{h}^{2}}+{{k}^{2}}=O{{C}^{2}}\] \[\Rightarrow \]               \[{{h}^{2}}+{{k}^{2}}={{(\sqrt{2})}^{2}}\] or\[{{x}^{2}}+{{y}^{2}}=2\]is required equation of locus of mid-point of chord subtending right angle at centre.