From a point Q, the length of the tangent to a circle is $24 c m$ and the distance of Q from the center is $25 c m .$ The radius of the circle is ___ $c m .$

Open in App

Solution

Given that $P Q = 24 c m$ and $O Q = 25 c m .$

To find OP, we apply Pythagoras Theorem in $△$ OPQ.

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

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

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

$\Rightarrow O P = 7 c m$

Suggest Corrections

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. 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

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

Join BYJU'S Learning Program

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, we apply Pythagoras Theorem in △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 $24 c m$ and the distance of Q from the center is $25 c m .$ The radius of the circle is ___ $c m .$

Given that $P Q = 24 c m$ and $O Q = 25 c m .$

To find OP, we apply Pythagoras Theorem in $△$ OPQ.

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

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

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

$\Rightarrow O P = 7 c m$