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

Given : A circle with center O and radius = 5cm 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 △AET 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 △BET, 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]

[From 3 and 4]

12