In Q.1 to 3, choose the correct option and give justification. 1. From a point , the length of the tangent to a circle is and the distance of from the centre is . The radius of the circle is (A) (B) (C) (D)

Understand the Question The tangent at any point of a circle is perpendicular to the radius through the point of contact, making . Triangle forms a right-angled triangle with hypotenuse and side . We can calculate the radius using the Pythagoras theorem: . Step 1 · Calculate the Radius using Pythagoras Theorem Let be the centre of the circle, be the given point, and be the point of contact of the tangent. Since the radius is perpendicular to the tangent at the point of contact, . In right , by Pythagoras theorem Answer > (A) Common Mistakes Identifying the Hypotenuse: The line connecting the centre to the external point () is opposite the angle and is therefore the hypotenuse, not the tangent (). Right Angle Placement: The right angle is formed at the point of contact (between the radius and tangent ), giving .

Circles | Exercise 10.2

Question 1

In Q.1 to 3, choose the correct option and give justification.

From a point $Q$ , the length of the tangent to a circle is $24\text{ cm}$ and the distance of $Q$ from the centre is $25\text{ cm}$ . The radius of the circle is (A) $7\text{ cm}$ (B) $12\text{ cm}$ (C) $15\text{ cm}$ (D) $24.5\text{ cm}$

Solution

Understand the Question

The tangent at any point of a circle is perpendicular to the radius through the point of contact, making $\angle \text{OPQ} = 90^\circ$ .
Triangle $\Delta \text{OPQ}$ forms a right-angled triangle with hypotenuse $\text{OQ} = 25\text{ cm}$ and side $\text{PQ} = 24\text{ cm}$ .
We can calculate the radius $\text{OP}$ using the Pythagoras theorem: $\text{OQ}^2 = \text{OP}^2 + \text{PQ}^2$ .

Step 1 · Calculate the Radius using Pythagoras Theorem

Let $O$ be the centre of the circle, $Q$ be the given point, and $P$ be the point of contact of the tangent. Diagram 1

Since the radius is perpendicular to the tangent at the point of contact, $\angle \text{OPQ} = 90^\circ$ .

In right $\Delta \text{OPQ}$ , by Pythagoras theorem

\begin{aligned} \text{OP}^2 + \text{PQ}^2 &= \text{OQ}^2 \\ \text{OP}^2 + 24^2 &= 25^2 \\ \text{OP}^2 + 576 &= 625 \\ \text{OP}^2 &= 625 - 576 \\ \text{OP}^2 &= 49 \\ \text{OP} &= \sqrt{49} = 7\text{ cm} \end{aligned}

Answer

(A) $7\text{ cm}$

Common Mistakes

Identifying the Hypotenuse: The line connecting the centre to the external point ( $OQ$ ) is opposite the $90^\circ$ angle and is therefore the hypotenuse, not the tangent ( $PQ$ ).
Right Angle Placement: The right angle is formed at the point of contact $P$ (between the radius $OP$ and tangent $PQ$ ), giving $\angle \text{OPQ} = 90^\circ$ .

Question 1

More questions in Exercise 10.2