In Q.1 to 3, choose the correct option and give justification. 2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that , then is equal to (A) (B) (C) (D)

Circles | Exercise 10.2

Question 2

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

In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that $\angle \text{POQ} = 110^\circ$ , then $\angle \text{PTQ}$ is equal to (A) $60^\circ$ (B) $70^\circ$ (C) $80^\circ$ (D) $90^\circ$

Solution

Setup: T is an external point. TP and TQ are two tangents from T to the circle, touching at P and Q respectively. OP and OQ are radii. Together O, P, T, Q form a quadrilateral OPTQ with four known or findable angles.

We will use the property that the radius is perpendicular to the tangent at the point of contact.

Step 1 — Identify angles

We are given that $\angle \text{POQ}$ is $110^\circ$ .

Tangent-Radius Perpendicularity Theorem: The radius to the point of tangency is always perpendicular to the tangent. So each radius-tangent pair forms a 90° angle.

We know that the radius is perpendicular to the tangent. So, $\text{OP} \perp \text{TP}$ . This means $\angle \text{OPT} = 90^\circ$ .

Also, $\text{OQ} \perp \text{TQ}$ . This means $\angle \text{OQT} = 90^\circ$ .

Step 2 — Calculate $\angle \text{PTQ}$

Consider the quadrilateral $\text{OPTQ}$ . Angle Sum of a Quadrilateral: The interior angles of any quadrilateral always add up to 360°. This follows from dividing the quadrilateral into two triangles, each having 180°.

The sum of angles in a quadrilateral is $360^\circ$ .

So, we can write the equation: $\angle \text{POQ} + \angle \text{OQT} + \angle \text{PTQ} + \angle \text{OPT} = 360^\circ$

Now, substitute the known angle values into the equation: $110^\circ + 90^\circ + \angle \text{PTQ} + 90^\circ = 360^\circ$

Combine the known angles: $290^\circ + \angle \text{PTQ} = 360^\circ$

Subtract $290^\circ$ from both sides to find $\angle \text{PTQ}$ : $\angle \text{PTQ} = 360^\circ - 290^\circ$

$\boxed{\angle \text{PTQ} = 70^\circ}$

Answer

(B) $70^\circ$

More questions in Exercise 10.2

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 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm

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

In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that $\angle \text{POQ} = 110^\circ$ , then $\angle \text{PTQ}$ is equal to (A) $60^\circ$ (B) $70^\circ$ (C) $80^\circ$ (D) $90^\circ$

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

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of $80^\circ$ , then $\angle \text{POA}$ is equal to (A) $50^\circ$ (B) $60^\circ$ (C) $70^\circ$ (D) $80^\circ$

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that

$\text{AB} + \text{CD} = \text{AD} + \text{BC}$

In Fig. 10.13, XY and $\text{X}'\text{Y}'$ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and $\text{X}'\text{Y}'$ at B. Prove that $\angle\text{AOB} = 90^\circ$ .

Q10

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Q11

Prove that the parallelogram circumscribing a circle is a rhombus.

Q12

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Q13

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.