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, PA=PBPA=PB
and OP is the bisector of ∠APB∠APB   [∵∵  two tangents to a circle from an external point are equally inclined to the line segment joining the centre to that point.
Also, OP⊥ABOP⊥AB and OP bisects AB at M  [∵∵ OP is the right bisector of AB]
∴∴ AM = MB = 9.629.62cm
=4.8cm.=4.8cm.
In right △AMO△AMO, we have
OA=6cmOA=6cm
and AM=4.8cm.AM=4.8cm.
∴∴  OM = OA2−AM2−−−−−−−−−−√OA2−AM2
=62−4.82−−−−−−−√=62−4.82
=12.96−−−−√=12.96
=3.6cm=3.6cm.
In right △PAO△PAO, we have
AP^{2 =} PM² + AM²
⇒x2=y2+(4.8)2⇒x2=y2+(4.8)2
⇒x2=y2+23.04⇒x2=y2+23.04...(i)
In right △PAO△PAO, we have
OP² = PA² + OA² [Note ∠PAO=90∘∠PAO=90∘, since AO is the radius at the point of contact]
⇒(y+3.6)2⇒(y+3.6)2
=x2+62=x2+62
⇒y2+7.2y+12.96⇒y2+7.2y+12.96
=x2+36=x2+36
⇒7.2y=46.08⇒7.2y=46.08 [using (i)]
⇒y=6.4cm⇒y=6.4cm
and
x² = (6.4)² + 23.04
= 40.96 + 23.04 = 64
⇒x=64−−√=8⇒x=64=8
∴∴ PA=8cm.