Question

7. In the given figure from an external point P, a tangent PT and a line segment PAB drawn to a circle
with centre O.ON is perpendicular on the chord AB. Prove that.
O PA PB = PN2 - AN2
(1) PN2-AN2 = Op2-OT2
(1) PA PB = PT2​

Answer 1

Answer:

Join OA

1. PA = (PN-AN) , PB = (PN+BN )

PA.PB = (PN-AN)(PN+BN)

ON Perpendicular AB so AN = BN

PA.PB = (PN-AN)(PN+AN) = PN^2-AN^2

2. PN^2-AN^2 =

IN ∆ONA

OA^2 = ON^2+AN^2

AN^2 = OA^2-ON^2

PN^2-(OA^2-ON^2)= PN^2+ON^2-OA^2

In ∆ ONP

OP^2 = ON^2+PN^2

so. OP^2- OA^2

OA= OT

PN^2-AN^2= OP^2-OT^2

3. from 1 & 2

PA.PB = OP^2-OT^2

IN ∆ OTP

OP^2 = PT^2+OT^2

OP^2-OT^2= PT^2

So PA.PB = PT^2