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

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 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

Solution

The radius, tangent, and line from the center to the external point form a right-angled triangle.

Step 1 — Set up the problem

Let's call the center of the circle O. Let the external point be Q. Let the point where the tangent touches the circle be P. We are given that the length of the tangent PQ is 24 cm. The distance from Q to the center OQ is 25 cm. We know that the radius OP is perpendicular to the tangent PQ. This means that ΔOPQ is a right-angled triangle at P.

Diagram 1

Step 2 — Apply Pythagoras theorem

Pythagoras Theorem: In a right-angled triangle, $\text{Hypotenuse}^2 = \text{Side}_1^2 + \text{Side}_2^2$ . Here OQ is the hypotenuse (side opposite the 90° angle at P), so $OQ^2 = OP^2 + PQ^2$ .

In the right-angled triangle ΔOPQ, we can use the Pythagoras theorem. The theorem states that $OP^2 + PQ^2 = OQ^2$ . Let's substitute the given values into the equation.

$OP^2 + 24^2 = 25^2$

$OP^2 + 576 = 625$

$OP^2 = 625 - 576$

$OP^2 = 49$

$OP = \sqrt{49}$

$\boxed{OP = 7 \text{ cm}}$

The radius of the circle is 7 cm.

Answer

(A) 7 cm

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.