A tangent at a point of a circle of radius meets a line through the centre at a point so that . Length is : (A) (B) (C) (D) .

Circles | Exercise 10.1

Question 3

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

(A) $12\text{ cm}$ (B) $13\text{ cm}$ (C) $8.5\text{ cm}$ (D) $\sqrt{119}\text{ cm}$ .

Solution

Understand the Question

The tangent at any point of a circle is perpendicular to the radius through the point of contact ( $OP \perp PQ$ ).
This forms a right-angled triangle $\triangle OPQ$ where $\angle OPQ = 90^\circ$ , making the line from the center $OQ$ the hypotenuse ( $12\text{ cm}$ ) and the radius $OP$ one of the legs ( $5\text{ cm}$ ).
We can calculate the length of the tangent $PQ$ by applying the Pythagoras theorem: $OQ^2 = OP^2 + PQ^2$ .

Step 1 · Apply Pythagoras Theorem in Right Triangle OPQ

Since the tangent at any point of a circle is perpendicular to the radius through the point of contact:

$OP \perp PQ \implies \angle OPQ = 90^\circ$ Diagram 1

In right-angled triangle $\triangle OPQ$ , by Pythagoras theorem

\begin{aligned} OQ^2 &= OP^2 + PQ^2 \\[0.6em] 12^2 &= 5^2 + PQ^2 \\[0.6em] 144 &= 25 + PQ^2 \\[0.6em] PQ^2 &= 144 - 25 \\[0.6em] PQ^2 &= 119 \\[0.6em] PQ &= \sqrt{119}\text{ cm} \end{aligned}

Answer

(D) $\sqrt{119}\text{ cm}$

Common Mistakes

Hypotenuse Misidentification: Assuming $PQ$ is the hypotenuse and calculating $\sqrt{12^2 + 5^2} = 13\text{ cm}$ (Option B). The right angle is at the point of contact $P$ , so $OQ = 12\text{ cm}$ is the hypotenuse.
Point of Tangency: Forgetting that perpendicularity is strictly between the radius $OP$ and tangent $PQ$ at point $P$ (i.e. $\angle OPQ = 90^\circ$ , not $\angle OQP$ ).

Question 3

More questions in Exercise 10.1