Q. 114.0( 14 Votes )

In figure, O is t

Answer :

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]



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