In Fig. 8, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle.
Solution
From the given figure, we have
TP = TQ (Two tangents, drawn from an external point to a circle, have equal length.)
and
∠TQO=∠TPO=90° (Tangent to a circle is perpendicular to the radius through the point of contact.)
In ∆TOQ,
QT2+OQ2=OT2
⇒QT2=132−52=144
⇒QT=12 cm
Now,
OT − OE = ET = 13 − 5 = 8 cm
Let QB = x cm.
∴ QB = EB = x (Two tangents, drawn from an external point to a circle, have equal length.)
Also,
∠OEB = 90° (Tangent to a circle is perpendicular to the radius through the point of contact.)
In ∆TEB,
EB2+ET2=TB2
⇒x2+82=(12−x)2
⇒x2+64=144+x2−24x
⇒24x=80
`=>x = 80/24=10/3`
`:.AB=2x=20/3cm`
Thus, the length of AB is `20/3 cm`