Question

Prove that length of tangent drawn from an external point of a circle are equal.

Answer 1

Hey...!!!
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Given: PT and TQ are two tangent drawn from an external point T to the circle C (O, r).

To prove: 1. PT = TQ

 2. ∠OTP = ∠OTQ 

Construction: Join OT.

Proof: We know that, a tangent to circle is perpendicular to the radius through the point of contact.

∴ ∠OPT = ∠OQT = 90°

In ΔOPT and ΔOQT,

OT = OT  (Common)

OP = OQ  ( Radius of the circle)

∠OPT = ∠OQT  (90°)

∴ ΔOPT=~ ΔOQT  (RHS congruence criterion)

⇒ PT = TQ  and ∠OTP = ∠OTQ (CPCT)

PT = TQ,

∴ The lengths of the tangents drawn from an external point to a circle are equal.

∠OTP = ∠OTQ,

∴ Centre lies on the bisector of the angle between the two tangents.

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I Hope it's help you...!!!

Answer 2

Hey mate..!!
Watch out the attachment...!

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Hope it helps..!

All the best for your exam ..!!

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