Prove that in two concentric , the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
Step-by-Step Solution
Key Concept: The radius drawn to the point of contact of a tangent is perpendicular to the tangent. In concentric circles the common centre lies on the perpendicular bisector of any chord of the larger circle. Hence the radius to the point of contact is the perpendicular bisector of the chord, proving the chord is bisected at that point.
1. Draw two concentric circles with common centre $O$. Let the larger circle have radius $R$ and the smaller circle have radius $r$ ($r
2. Let $AB$ be a chord of the larger circle which touches the smaller circle at $T$.
3. Since $T$ is the point of contact, $OT$ is a radius of the smaller circle. By the tangent‑radius theorem, $OT\perp AB$.
4. In triangle $\triangle OAB$, we have $OA=OB=R$ (radii of the larger circle). Hence $O$ is equidistant from $A$ and $B$, so the line through $O$ perpendicular to $AB$ must be the perpendicular bisector of $AB$.
5. The line $OT$ is perpendicular to $AB$ and passes through $O$; therefore $OT$ is the perpendicular bisector of $AB$. Consequently, the foot of the perpendicular, which is the point $T$, divides $AB$ into two equal parts: $AT = TB$.
6. Hence the chord $AB$ of the larger circle is bisected at the point of contact $T$ with the smaller circle.
Correct Answer: The chord $AB$ of the larger circle is bisected at the point $T$ where it touches the smaller circle; i.e., $AT = TB$.