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Question

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 length of AB.

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Solution

Clearly OPT=90o

Applying Pythagoras in OPT, we have

OT2=OP2+PT2

132=52+PT2

PT2=16925=144

PT=12 cm

Since lengths of tangents drawn from a point to a circle are equal. Therefore,

AP=AE=x(say)

AT=PTAP=(12x)cm

Since AB is the tangent to the circleE. Therefore, OEAB

OEA=90o

AET=90o

AT2=AE2+ET2 [Applying Pythagoras Theorem in AET]

(12x)2=x2+(135)2

14424x+x2=x2+64

24x=80

x=103cm

Similarly, BE=103cm

AB=AE+BE=(103+103)cm=203cm

1029519_1009695_ans_8349c848016648469d36d72d57d210d9.png

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