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: Understanding the Question:
The problem involves two concentric circles sharing a common center \(O\).
The radius of the larger outer circle is given as 25 cm.
A chord \(PQ\) of the outer circle touches the inner circle at point \(R\), meaning \(PQ\) is a tangent to the inner circle at \(R\).
The length of the chord \(PQ\) is 14 cm. We need to determine the radius of the inner circle.

Step 2: Key Formula or Approach:
1. A line drawn from the center of a circle perpendicular to a chord bisects the chord.
2. The radius of a circle is always perpendicular to the tangent at the point of contact.
3. Let \(r\) be the radius of the inner circle (\(OR\)) and \(R\) be the radius of the outer circle (\(OP\)).
4. Since \(OR \perp PQ\), \(\Delta ORP\) is a right-angled triangle, and we can apply the Pythagorean theorem:
\[ OP^2 = OR^2 + PR^2 \]

Step 3: Detailed Explanation:
1. Let \(O\) be the common center. Let \(OP\) be the radius of the outer circle, so \(OP = R = 25\ \text{cm}\).
2. Since \(PQ\) is a chord of the outer circle and is tangent to the inner circle at \(R\), the radius \(OR\) is perpendicular to \(PQ\).
\[ OR \perp PQ \]
3. The perpendicular from the center to a chord bisects the chord. Therefore, \(R\) is the midpoint of \(PQ\).
\[ PR = RQ = \frac{PQ}{2} = \frac{14}{2} = 7\ \text{cm} \]
4. In the right-angled triangle \(ORP\), apply Pythagoras' theorem:
\[ OP^2 = OR^2 + PR^2 \]
\[ 25^2 = r^2 + 7^2 \]
\[ 625 = r^2 + 49 \]
\[ r^2 = 625 - 49 \]
\[ r^2 = 576 \]
\[ r = \sqrt{576} = 24\ \text{cm} \]
5. Thus, the radius of the inner circle is 24 cm.

Step 4: Final Answer:
The radius of the inner circle is 24 cm, which corresponds to option (B).