Chapter 10 · Question 5
Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.
Q5
Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.
Answer Revealed
Direct Answer:
Let two concentric circles (larger) and (smaller) have common centre . Let chord of touch at point . Join . Since is a tangent to at , by Theorem 10.1, . Now is the perpendicular from the centre to the chord of . The perpendicular from the centre of a circle to a chord bisects the chord. Therefore, .
Simple Explanation
The chord of the larger circle that touches the smaller circle gets cut exactly in half at the point where it touches. This happens because the line from the centre to that touching point is perpendicular to the chord, and a perpendicular from the centre to any chord always bisects it.
Exam-Ready Structure
Given: Two concentric circles (outer) and (inner) with common centre . A chord of touches at point . To Prove: (the chord is bisected at ). Construction: Join . Proof: Since is a tangent to the smaller circle at point , and is the radius of through , by Theorem 10.1, . Now, in the larger circle , is a chord and is the perpendicular drawn from the centre to this chord. A property from Class IX states: The perpendicular from the centre of a circle to a chord bisects the chord. Therefore, . Hence, the chord of the larger circle is bisected at the point of contact with the smaller circle.
Key Points
- Two concentric circles: common centre O
- Chord AB of larger circle touches smaller circle at P
- OP ⟂ AB (Theorem 10.1, tangent to smaller circle)
- Perpendicular from centre to chord bisects the chord
- Therefore AP = BP