A) \[PM=MQ\]
B) \[PQ\bot AB\]
C) Both (A) and (B)
D) \[PQ=AB\]
Correct Answer: C
Solution :
In the given figure, PQ is the common chord of the two circles. \[\Rightarrow \] AB bisects the common chord PQ at M. \[\therefore \] PM = MQ Moreover, PQ is perpendicular to AB. \[\therefore \] Option (C) is correct.You need to login to perform this action.
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