Q) PT is the tangent to the circle centered at O. OC is Perpendicular to the chord AB. Prove that PA.PB = PC2-AC2. Ans: Let’s starts from LHS: PA . PB = (PC – AC) (PC + BC) Given that PC is to chord AB, therefore it bisects chord AB (by circle’s property) Hence, AC = […]

In the given figure, PT is the tangent to the circle centered at O. OC

Q) PT is the tangent to the circle centered at O. OC is Perpendicular to the chord AB. Prove that PA.PB = PC²-AC².

Ans:

Let’s starts from LHS:

PA . PB = (PC – AC) (PC + BC)

Given that PC is $\perp$ to chord AB, therefore it bisects chord AB (by circle’s property)

Hence, AC = BC

Therefore LHS:

PA. PB = (PC – AC) (PC + AC)

= PC²– AC² = RHS

Hence Proved