Question

In two concentric circles with centre O, the radius of outer circle is 25 cm. Chord PQ of the outer circle is tangent to the inner circle at R. If PQ = 14 cm, then the radius of the inner circle is :

• A. \(\sqrt{429}\) cm
• B. 24 cm
• C. \(\sqrt{674}\) cm
• D. 20 cm

Hint:

For concentric circle problems where a chord of the outer circle touches the inner circle, always construct the right-angled triangle formed by the inner radius \(r\), half the chord length \(a\), and the outer radius \(R\).
The relation is always:
\[ R^2 = r^2 + \left(\frac{\text{Chord}}{2}\right)^2 \]
Remembering common Pythagorean triples like \((7, 24, 25)\) helps you find the answer immediately without doing long calculations!

Answer 1

The Correct Option is B

Step 1: Understand what is being asked.
Two circles share centre $O$. The outer circle has radius 25 cm, and a chord $PQ=14$ cm of the outer circle touches the inner circle at $R$. We need the inner radius. Let us use the standard relation between a chord of a circle and its perpendicular distance from the centre, since the inner radius is exactly that perpendicular distance here.

Step 2: Recognise the inner radius as a distance from centre to chord.
Because $PQ$ is tangent to the inner circle at $R$, the segment $OR$ is perpendicular to $PQ$ (a tangent is always perpendicular to the radius at the point of contact). So $OR$ is nothing but the perpendicular distance of the centre $O$ from the chord $PQ$ of the outer circle. Call this distance $d$, so $r=d=OR$ is what we want.

Step 3: Apply the chord-distance formula.
For any chord of length $l$ in a circle of radius $R$, the perpendicular distance $d$ from the centre satisfies:
\[ d = \sqrt{R^2 - \left(\frac{l}{2}\right)^2} \]
Here $R=25$ cm (outer radius) and $l=14$ cm (chord length), so half the chord is 7 cm:
\[ d = \sqrt{25^2 - 7^2} = \sqrt{625-49} = \sqrt{576} \]
Step 4: Evaluate the square root.
\[ d = 24 \text{ cm} \]
Final Answer:
The radius of the inner circle is 24 cm, matching option (B).
\[ \boxed{24 \text{ cm}} \]