AB is a chord of length …

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.
CONSTRUCTION :  Join OP and OA. Let OP and AB intersect at M.
Let PA = x cm and PM = y cm.
Now, {tex}PA = PB{/tex}
and OP is the bisector of {tex}\angle A P B{/tex}   [{tex}\because{/tex}  two tangents to a circle from an external point are equally inclined to the line segment joining the centre to that point.
Also, {tex}O P \perp A B{/tex} and OP bisects AB at M  [{tex}\because{/tex} OP is the right bisector of AB]
{tex}\therefore{/tex} AM = MB = {tex}\frac { 9.6 } { 2 }{/tex}cm
{tex}= 4.8 cm.{/tex}
In right {tex}\triangle A M O{/tex}, we have
{tex}OA = 6cm{/tex}
and {tex}AM = 4.8cm.{/tex}
{tex}\therefore{/tex}  OM = {tex}\sqrt { O A ^ { 2 } - A M ^ { 2 } }{/tex}
{tex}= \sqrt { 6 ^ { 2 } - 4.8 ^ { 2 } }{/tex}
{tex}= \sqrt { 12.96 }{/tex}
{tex}= 3.6 \mathrm { cm }{/tex}.
In right {tex}\triangle P A O{/tex}, we have
AP^{2 =} PM² + AM²
{tex}\Rightarrow x ^ { 2 } = y ^ { 2 } + ( 4.8 ) ^ { 2 }{/tex}
{tex}\Rightarrow \quad x ^ { 2 } = y ^ { 2 } + 23.04{/tex}...(i)
In right {tex}\triangle P A O{/tex}, we have
OP² = PA² + OA² [Note {tex}\angle P A O = 90 ^ { \circ }{/tex}, since AO is the radius at the point of contact]
{tex}\Rightarrow \quad ( y + 3.6 ) ^ { 2 }{/tex}
{tex}= x ^ { 2 } + 6 ^ { 2 }{/tex}
{tex}\Rightarrow \quad y ^ { 2 } + 7.2 y + 12.96{/tex}
{tex}= x ^ { 2 } + 36{/tex}
{tex}\Rightarrow 7.2 y = 46.08{/tex} [using (i)]
{tex}\Rightarrow \quad y = 6.4 \mathrm { cm }{/tex}
and
x² = (6.4)² + 23.04
= 40.96 + 23.04 = 64
{tex}\Rightarrow x = \sqrt { 64 } = 8{/tex}
{tex}\therefore{/tex} {tex}PA = 8cm.{/tex}