Question

A tangent PQ at a point P of a circle of radius 10 cm meets a line through the centre O at a point Q so that OQ = 12 cm. The length of PQ will be:

• A. 12 cm
• B. 13 cm
• C. $2\sqrt{11}$ cm
• D. $3\sqrt{5}$ cm

Hint:

For tangents, always use \( PQ^2 = OQ^2 - OP^2 \). The tangent, radius, and line through the centre form a right-angled triangle.

Answer 1

The Correct Option is D

Step 1: Concept used.
A tangent to a circle is always perpendicular to the radius at the point of contact. Thus, $\triangle OPQ$ is a right-angled triangle at \( P \).

Step 2: Apply Pythagoras theorem.
\[ OQ^2 = OP^2 + PQ^2 \] \[ PQ^2 = OQ^2 - OP^2 = 12^2 - 10^2 = 144 - 100 = 44 \] \[ PQ = \sqrt{44} = 2\sqrt{11} \text{ cm} \]
Step 3: Verify the options.
Hence, \( PQ = 2\sqrt{11} \) cm, which is not in simplified radical form equal to \( 3\sqrt{5} \). Wait — since \( \sqrt{44} = 2\sqrt{11} \), option (C) is correct, not (D).
Step 4: Conclusion.
The correct length of \( PQ \) is \( \boxed{2\sqrt{11}\ \text{cm}} \).