Given, OT =13cm and OP =5cm
Since, if we drawn a line from the centre to the tangent of the circle. It is always perpendicular to the tangent i.e., `OPbotPT.`
In right angled `DeltaOPT, OT^(2)=OP^(2)+PT^(2)`
[by Pythagoras theorem, `("hypotenuse")^(2)=("base")^(2)+("perpendicular")^(2)`]
`impliesPT^(2)=(13)^(2)-(5)^(2)=169-25=144`
`impliesPT=12cm`
Since, the langth of pair of tangents from an external point T is equal.
`:.QT=12cm`
Now, TA=PT-PA
`impliesTA=12-PA.....(i)`
and TB=QT-QB
`impliesTB=12-QB......(ii)`
Again using the property, length of pair of tanagets from an external point is equal.
`:.PA=AEand QB=EB.......(iii)`
`:.OT=13cm`
`:.ET=OT-OE" "[:.OE=5cm="radius"]`
`impliesET=13-5`
`impliesET=8cm`
Since, AB is a tangent and OE is the radius.
`:.OEbotAB`
`impliesangleOEA=90^(@)`
`:.angleAET=180^(@)-angleOEA` [linear pair]
`impliesangleAET=90^(@)`
Now, in right angled `DeltaAET`,
`(AT)^(2)=(AE)^(2)+(ET)^(2)` [by Pythagoras theorem]
`implies(PT-PA)^(2)=(AE)^(2)+(8)^(2)`
`implies(12-PA)^(2)=(AP)^(2)+(8)^(2)` [from Eq. (iii)]
`implies144+(PA)^(2)-24.PA=(PA)^(2)+64`
`implies24.PA=80`
`impliesPA=(10)/(3)cm`
`:.AE=(10)/(3)cm` [from Eq. (iii)]
Join OQ.
Similarly `BE=(10)/(3)cm`
Hence, `AB=AE+EB`
`=(10)/(3)+(10)/(3)`
`=(20)/(3)cm`
Hence, the required length AB is `=(20)/(3)cm`.
