Question

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.

Answer 1

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 pythagoras 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