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,

QT^{2}+OQ^{2}=OT^{2}

⇒QT^{2}=13^{2}−5^{2}=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,

EB^{2}+ET^{2}=TB^{2}

⇒x^{2}+8^{2}=(12−x)^{2}

⇒x^{2}+64=144+x^{2}−24x

⇒24x=80

`=>x = 80/24=10/3`

`:.AB=2x=20/3cm`

Thus, the length of AB is `20/3 cm`