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.You need to login to perform this action.
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