Given: Two tangents PA and PB to a circle with centre O, from an exterior point P.
R.T.P : PA =
Proof: In △OAP; ∠OAP = 90°
∴ AP2 = OP2 - O2
[∵ Square of the hypotenuse is equal to the sum of squares on the other two sides - Pythagaros theorem]
[∵ O = OB, radii of the same circle]
= BP2 [∵ In AOBP; OB2 + BP2 = OP2] [BP2 - OP2 - 2]
AP2 - 2
PA - Hence proved