Question

In figure,from an external point $P$,a tangent $PT$ and a line segment $PAB$ is drawn to a circle with centre $O$.$ON$ is perpendicular on the chord $AB$.
Prove that:-

$\left(i\right)PA.PB={PA}^2-{AN}^2$

$\left(ii\right){PN}^2-{AN}^2={OP}^2-{OT}^2$

$\left(iii\right)PA.PB={PT}^2$

Answer 1

$\to \left(i\right)$

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

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

[$AN=BN$ as $ON\perp AB$ and so bisect it]

$PA.PB=P{N}^2-A{N}^2$ $\left[\right(a+b\left)\right(a-b)=a^2-b^2]$

$\left(ii\right)P{N}^2-A{N}^2=(O{P}^2-O{N}^2)-A{N}^2$

[ in $\Delta ONA,O{A}^2=O{N}^2+A{N}^2,$ (By Pythagoras theorem)]

$P{N}^2-A{N}^2=O{P}^2-O{T}^2$ [ $OA=OT=$ radii of circle]

(iii) From parts (i) & (ii),

$PA.PB=O{P}^2-O{T}^2$

$PA.PB=P{T}^2$

[In $\Delta OTP,P{T}^2=O{P}^2-O{T}^2$ (By Pythagoras theorem)]

$PA.PB=P{T}^2$

Hence proved.