In the adjoining figure, AB is a chord of length 9.6 cm of a circle with centre O and radius 6 cm. The tangents at A and B intersect at P. Find the length of PA.

Explanation

GIVEN A circle with centre `O` and radius `6cm`. `AB` is a chord of length `9.6cm`. The tangents at `A` and `B` intersect at `P`.
TO FIND The length `PA`.
CONSTRUCTION Join `OP` and `AB` intersect at `M`.
Let `PA=xcm ` and `PM=y cm`.
Now, `PA=PB` [`:'` tangents from an external point are equal and `OP` is the bisector of `/_APB`
[`:'` two tangents to a circle from an external point are equally inclined to the line segment joining the centre the that point].
Also, `OPbotAB` and `OP` bisects `AB` at `M`
[`:' OP` is right bisector of `AB`].
`:. AM=MB=(9.6)/(2)cm=4.8cm`
In right `DeltaAMO`, we have `OA=6cm` and `AM=4.8cm`
`:.OM=sqrt(OA^(2)-AM^(2))=sqrt(6^(2)-4.8^(2))=sqrt(12.96)=3.6cm`
In right `DeltaAMP`, we have
`AP^(2)=PM^(2)+AM^(2)impliesx^(2)=y^(2)+(4.8)^(2)`
`impliesx^(2)=y^(2)+23.04`..........`(i)`
In right DeltaPAO`, we have
`OP^(2)=PA^(2)+OA^(2)` [Note `/_PAO=90^(@)`, since `AO` is the radius at the point of contact]
`implies(y+3.6)^(2)=x^(2)+6^(2)` [`:' OP=PM+MO=(y+3.6)cm`]
`impliesy^(2)+7.2y+12.96=x^(2)+36implies7.2y=46.08` [using `(i)` ]
`impliesy=6.4cm`
Putting this value of `y` in `(i)`, we get
`x^(2)=(6.4)^(2)+23.04=40.96+23.04=64impliesx=sqrt(64)=8`.
`:.PA=8cm`