Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.
Step-by-Step Solution
Key Concept: Let $AB$ be chord of outer circle touching inner circle at $P$. $OP \perp AB$ (radius $\perp$ tangent). Perpendicular from centre to a chord bisects the chord $\Rightarrow AP = PB$.
Let $AB$ touch smaller circle at $P$. Since $OP$ is radius of smaller circle, $OP \perp AB$. [1.0 Mark]
Since $AB$ is chord of larger circle and $OP \perp AB$, the perpendicular from centre bisects the chord $\Rightarrow AP = PB$. Proved! [1.0 Mark]
---
🎯 Official CBSE Marking Scheme:
Stating $OP \perp AB$ (radius $\perp$ tangent): 1.0 Mark
Applying perpendicular from centre bisects chord $\Rightarrow AP = PB$: 1.0 Mark
Correct Answer: