$$C_1$$ and $$C_2$$ are two circles such that $$C_1$$ lies inside $$C_2$$. $$AB$$ is a chord of circle $$C_2$$ touching $$C_1$$ at $$P$$.

Solution

The radius of $$C_1$$ to $$P$$ is perpendicular to the tangent chord $$AB$$. If the circles are concentric, this is the perpendicular from the centre of $$C_2$$ to its chord, which bisects the chord, so $$AP = PB$$. In general the centre of $$C_2$$ is not on this perpendicular, so options a, b and d need not hold.

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