O is the centre of a circle with radius 5 centimetre if T is a point such that OT equals to 13 cm find the length of AB.
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Answer:
Step-by-step explanation:
Clearly ∠OPT=90
o
Applying Pythagoras in △OPT, we have
OT
2
=OP
2
+PT
2
⇒13
2
=5
2
+PT
2
⇒PT
2
=169−25=144
⇒PT=12 cm
Since lengths of tangents drawn from a point to a circle are equal. Therefore,
AP=AE=x(say)
⇒AT=PT−AP=(12−x)cm
Since AB is the tangent to the circleE. Therefore, OE⊥AB
⇒∠OEA=90
o
⇒∠AET=90
o
⇒AT
2
=AE
2
+ET
2
[Applying Pythagoras Theorem in △AET]
⇒(12−x)
2
=x
2
+(13−5)
2
⇒144−24x+x
2
=x
2
+64
⇒24x=80
⇒x=
3
10
cm
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