Question

A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is:
$
  {\text{A}}{\text{. 12 cm}} \\
  {\text{B}}{\text{. 13 cm}} \\
  {\text{C}}{\text{. 8}}{\text{.5 cm}} \\
  {\text{D}}{\text{. }}\sqrt {113} {\text{ cm}} \\
$

Answer 1

Hint: First we have to draw a suitable diagram for this type of question so that question becomes a little bit easy. The tangent PQ and radius of the circle subtend a right angle. Use Pythagoras’ Theorem. Because this is the easiest way of solving this type of question.

Complete step-by-step answer:

Here $\Delta OPQ$ is a right angled triangle. And we know that we can apply Pythagoras theorem in the right angled triangle.
Pythagoras theorem :- in a right angled triangle, square of hypotenuse is equal to sum of square of perpendicular and base that is ${{\text{H}}^2} = {{\text{P}}^2} + {{\text{B}}^2}$
Applying Pythagoras’ Theorem in $\Delta OPQ$
Here hypotenuse = OQ
Base = OP
Perpendicular = PQ
So from Pythagoras theorem we can write,
$
  O{Q^2} = O{P^2} + P{Q^2} \\
  P{Q^2} = {12^2} - {5^2} \\
  P{Q^2} = 113 \\
  PQ = \sqrt {113{\text{ }}} {\text{ cm}} \\
$
The correct option is (D).


Note: In these types of questions, Pythagoras’ theorem and the property of a tangent to a circle are important. So remember these properties for further this type of question. We should also remember that tangent always makes a right angle with radius.