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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


Answer :




Since QT is a tangent to the circle at T and OT is radius,

\(\therefore \ OT ⊥ QT \)

It is given that OQ = 25 cm and QT = 24 cm,

\(\therefore \) By Pythagoras theorem, we have

\( \Rightarrow OQ^2 \ = \ QT^2 \ + \ OT^2 \)
\(\Rightarrow OT^2 \ = \ OQ^2 \ - \ QT^2 \)
\( \Rightarrow \ OT^2 \ = \ 25^2 \ - \ 24^2 \)
\( \Rightarrow OT^2 \ = \ 49 \)

\( \Rightarrow \ OT \ = \ \sqrt{49} \ = \ 7 \)

\(\therefore \) radius of the circle is 7 cm

\(\therefore \) Option (A) is correct.

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