Question

prove that the length of the tangents drawn from an external point to a circle are equal, hence show that the centre lies on the bisector of the angle between the two tangents?

Answer 1

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.

Answer 2

Given: A circle with centre O. Two tangenta PA and PB drawn from an extenal point P touching the circle at point A and B respectively. OA and OB are the radii.                                                                                                                          (1/2 marks)

To prove: PA = PB.                                                                                                                                                            (1/2 marks)

Construction: Join OA, OB and OP.                                                                                                                               (1/2 marks)

(1/2 marks for the diagram)

Proof: PAO and PBO are two triangles.

In triangles PAO and PBO,

OA = OB (Radii of same circle)

|PAO = |PBO  (Angle between radii and tangents)

OP = OP (Common side)

Therefore, Triangle PAO ~= Triangle PBO. (R.H.S. congruence)

=> PA = PB. (CPCT)