From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. The radius of the circle is
___cm.

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Solution

Given that PQ = 24 cm and OQ = 25 cm.

To find OP, apply Pythagoras theorem for $△$ OPQ.

${O P}^{2} + {Q P}^{2} = {O Q}^{2}$

${O P}^{2} + 24^{2} = 25^{2}$

${O P}^{2} = 625 - 576 = 49$

$\Rightarrow O P = 7 cm$

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Similar questions

The length of a tangent from a point Q to a circle is 24 cm. The distance between Q and the center of the circle is 25 cm. The radius of the circle is _____ cm.

Q. Question 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

Q. Find the length of chord of a circle whose radius is

25 c m

and The distance from the center is

24 c m

Q. 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 c m

. The radius of the circle is

A point P is 25 cm away from the centre of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.

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From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. The radius of the circle is ___cm.

Given that PQ = 24 cm and OQ = 25 cm. To find OP, apply Pythagoras theorem for △OPQ. OP2+QP2=OQ2 OP2+242=252 OP2=625−576=49 ⇒OP=7 cm

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. The radius of the circle is
___cm.

Given that PQ = 24 cm and OQ = 25 cm.

To find OP, apply Pythagoras theorem for $△$ OPQ.

${O P}^{2} + {Q P}^{2} = {O Q}^{2}$

${O P}^{2} + 24^{2} = 25^{2}$

${O P}^{2} = 625 - 576 = 49$

$\Rightarrow O P = 7 cm$