O is the centre of a circle of radius 5cm. T is a point such that OT=13cm and OT intersects the circle at E, find the length AB.

Explanation

Since, `angleOPT=90^(@)`
(radius through point of contact is `_|_` to the tangent)
`:.` In right `triangleOPT,`
`OP^(2)+PT^(2)=OT^(2)" "`(by Pythagoras theorem)
`implies" "PT^(2)=OT^(2)-OP^(2)`
`implies" "PT^(2)=(13)^(2)-(5)^(2)=(12)^(2)`
`implies" "PT=12cm`
Let`AP=xcm`
`:." "AE=AP=x" "`(lengths of tangents from an external point are equal)
`:." "AT=TP-AP=12-x`
`ET=OT-OE=13-5=8cm`
Now, since `angleAEO=90^(@)" "`(radius through point of contact is `_|_` to the tangent)
`:." "angleAET=90^(@)" "`(L.P.A)
`:.` In right `triangleAET,` by Pythagoras theorem,
`AE^(2)+ET^(2)=AT^(2)`
Circles
`implies" "x^(2)+(8)^(2)=(12-x)^(2)`
`implies" "x^(2)+64=144+x^(2)-24x`
`implies" "24x=144-64=80`
`implies" "x=(80)/(24)=(10)/(3)`
Similarly,`" "BE=(10)/(3)cm`
`:." "AB=AE+BE=((10)/(3)+(10)/(3))cm=(20)/(3)cm`
Hence,`" "AB=(20)/(3)cm`