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.You need to login to perform this action.
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