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Circles

Question
CBSEENMA10009660

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2 ∠OPQ.

Solution


Given: TP and TQ are two tangents of a circle with centre O and P and Q are points of contact
To prove: ∠PTQ=2∠OPQ
Suppose ∠PTQ=θ.
Now by theorem, "The lengths of tangents drawn from an external point to a circle are equal".
So, TPQ is an isosceles triangle.
Therefore, ∠TPQ=∠TQP= 1 half left parenthesis 180 to the power of straight o minus straight theta right parenthesis equals space 90 to the power of straight o space minus straight theta over 2 space
Also by theorem "The tangents at any point of a circle is perpendicular to the radius through the point of contact" ∠OPT=90°.
Therefore,
∠OPQ=∠OPT−∠TPQ
equals 90 to the power of straight o space minus open parentheses 90 to the power of straight o space minus 1 half straight theta close parentheses
space equals space 1 half straight theta
space equals space 1 half angle PTQ
Hence, ∠PTQ=2∠OPQ.

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