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
