Question

If $PAB$ is a secant to a circle of centre $O$ intersecting the circle of $A$ and $B$ and $PT$ is tangent segment, then prove that-
$PA\times PB=P{T}^2$

Answer 1

If $PAB$ is a secant to a circle intersecting it at A and B and PT is a tangent then $PA.PB={PT}^2$

Here $PAB$ is secant intersecting the circle with centre O at $A$ and $B$ and a tangent $PT$ at $T$

We make construction that is $OM\perp AB$ is drawn $OA,OP,OT$ are joined.

Now $PA=PM–AM$

$PB=PM+MB$

As we know that $AM=BM$ [perpendicular drawn from the centre of the circle to a chord is also a bisector of the chord]

$PA.PB=(PM-AM).(PM+AM)$

$PA.PB={PM}^2-{AM}^2$

Also $OM\perp AB$

we can apply Pythagoras theorem in $\Delta OMP$

${PM}^2={OP}^2-{OM}^2$

Now apply Pythagoras theorem in $\Delta OMA$

${AM}^2={OA}^2-{OM}^2$

now this values in the above equation so we get

$PA.PB={PM}^2-{AM}^2$

$PA.PB={(OP}^2-{OM}^2)-{(OA}^2-{OM}^2)$

$PA.PB={OP}^2-{OM}^2-{OA}^2+{OM}^2$

$PA.PB={OP}^2-{OA}^2$

$PA.PB={OP}^2-{OT}^2$ as $OA=OT$ (radii)

As radius is perpendicular to the tangent so this will form a right angled triangle

We can apply Pythagoras theorem in $\Delta OPT$

We get $P{T}^2=O{P}^2-O{T}^2$

By putting this value in above equation we get $PA.PB={PT}^2$

Hence proved