Question

Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.

Hint:

Remember: The line drawn from the center of a circle to the point of contact of a tangent is always perpendicular to the tangent.

Answer 1

Step 1: Given.
Two concentric circles (same center \( O \)) are given. Let the larger circle have radius \( R \) and the smaller circle have radius \( r \).
A chord \( AB \) of the larger circle touches the smaller circle at point \( P \).

Step 2: To Prove.
We have to prove that \( OP \) bisects the chord \( AB \), i.e., \( AP = PB \).

Step 3: Construction and reasoning.
Draw radii \( OA \) and \( OB \) to the ends of the chord. Since \( OP \) is perpendicular to the chord at the point of contact \( P \) (a property of tangent and radius), \[ OP \perp AB \] Thus, \( OP \) bisects the chord \( AB \).
Step 4: Proof using geometry.
In triangles \( OAP \) and \( OBP \): \[ OA = OB \quad (\text{Radii of the same circle}) \] \[ OP = OP \quad (\text{Common side}) \] \[ \angle OAP = \angle OBP = 90^\circ \] Therefore, by RHS congruence, \[ \triangle OAP \cong \triangle OBP \Rightarrow AP = PB \] Step 5: Conclusion.
Hence proved that the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.