In figure. O is the centre of a circle of radius 5 cm, T is a point such that OT = 13 cm and OT intersects the circle at E. If AB is the tangent to the circle at E, find the length of AB.

#### Solution

**Given:** A circle with center O and radius = 5 cm T is a point, OT = 13 cm. OT intersects the circle at E and AB is the tangent to the circle at E.

**To Find:** Length of AB

OP ⏊ PT ......[Tangent at a point on the circle is perpendicular to the radius through point of contact]

By phythagoras theorem in △OPT right angled at P

(OT)^{2} = (OP)^{2}+ (PT)^{2}

(13)^{2} = (5)^{2} + (PT)^{2}

(PT)^{2}= 169 - 25 = 144

PT = 12 cm

PT = TQ = 12 cm ......[Tangents drawn from an external point to a circle are equal]

Now, OT = OE + ET

ET = OT - OE = 13 - 5 = 8 cm

Now, as Tangents drawn from an external point to a circle are equal.

AE = PA .....[1]

EB = BQ ......[2]

Also OE ⏊ AB ......[Tangent at a point on the circle is perpendicular to the radius through point of contact]

∠AEO = 90°

∠AEO + ∠AET = 180° .......[By linear Pair]

∠AET = 90°

In ΔAET By Pythagoras Theorem

(AT)^{2} = (AE)^{2} + (ET)^{2 }.......[Here AE = PA as tangents drawn from an external point to a circle are equal]

(PT - PA)^{2} = (PA)^{2} + (ET)^{2}

(12 - PA)^{2} = (PA)^{2} + (8)^{2} ......[From 1]

144 + (PA)^{2} - 24PA = (PA)^{2} + 64

24PA = 80 .......[3]

∠AET + ∠BET = 180 .......[Linear Pair]

90 + ∠BET = 180

∠BET = 90

In ΔBET, By Pythagoras Theorem

(BT)^{2} = (BE)^{2}+ (ET)^{2}

(TQ - BQ)^{2} = (BQ)^{2} + (ET)^{2} ......[From 2]

(12 - BQ)^{2} = (BQ)^{2} + (8)^{2}

144 + (BQ)^{2} - 24BQ = (BQ)^{2} + 64

24BQ = 80 ......[4]

So, AB = AE + BE

AB = PA + BQ .......[From 1 and 2]

AB = `10/3 + 10/3` ......[From 3 and 4]

AB = `20/3` cm