Question

$\large \pink{\underbrace{\frak{ \red{Question}}}}$

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. Find the radius of the circle.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

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.

Let assume that the center of circle be O and radius is r cm.

Let tangent drawn from a point Q touches the circle at P.

• So, PQ = 24 cm

Now, Q is at a distance of 25 cm from centre.

• So, OQ = 25 cm

Construction :- Join OP

• So, OP = r cm

As OP is radius and QP is a tangent drawn from external point Q

So, $\rm\implies \:OP\: \perp \:PQ$

Now, In right-angle $\triangle$ OPQ

By using Pythagoras Theorem, we have

$\rm \: {OQ}^{2} = {OP}^{2} + {PQ}^{2}$

$\rm \: {25}^{2} = {r}^{2} + {24}^{2}$

$\rm \: 625 = {r}^{2} + 576$

$\rm \: {r}^{2} = 625 - 576$

$\rm \: {r}^{2} = 49$

$\rm \: {r}^{2} = {7}^{2}$

$\bf\implies \:r \: = \: 7 \: cm$

$\rule{190pt}{2pt}$

Additional Information :-

1. Length of tangents drawn from external point are equal.

2. Only two tangents can be drawn to a circle from external point.

3. A circle can have infinitely many tangents.

Answer 2

Answer:

7cm

Step-by-step explanation:

Let QP be the tangent, such that, Point of contact is P.

Length of the tangent to a circle = 24cm

$$PQ=24cm$$

Let O be the centre of the circle.

OQ=25cm

We have to find the radius OP

Since QP is tangent

OP perpendicular to QP (Since, Tangent is Perpendicular to Radius at the point of contact)

So, ∠OPQ=90°

So apply Pythogoras theorem to right triangle, OPQ;

$OP^{2} = OQ^{2} - PQ^{2}$

$OP^{2} = 25^{2} - 24^{2}$

$OP^{2} = 49cm$

$OP = \sqrt{49}$

$OP = 7cm$

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