How i solve the problem from class 10th ncert of exercise 10.2 of circles chapter of question no.1, i think this an important question for class 10th, give me the best way to solve this problem 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

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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 centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm Q.1

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First, draw a perpendicular from the center O of the triangle to a point P on the circle which is touching the tangent. This line will be perpendicular to the tangent of the circle.

So, OP is perpendicular to PQ i.e. OP âŠ¥ PQ

From the above figure, it is also seen that â–³OPQ is a right angled triangle.

It is given that

OQ = 25 cm and PQ = 24 cm

By using Pythagoras theorem in â–³OPQ,

OQ

^{2}= OP^{2}+PQ^{2}(25)

^{2 }= OP^{2}+(24)^{2}OP

^{2}= 625-576OP

^{2}= 49OP = 7 cm

So, option A i.e. 7 cm is the radius of the given circle.