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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.

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Answer \frac{20}{3}\, cm
Solution
  Given: Radius = 5 cm, OT = 13 cm
OP\perp PT         [  PT is tangent]
Using Pythagoras theorem \bigtriangleupOPT
H^{2}+B^{2}+P^{2}
\left ( OT \right )^{2}= \left ( OP \right )^{2}+\left ( PT \right )^{2}
\left ( 13 \right )^{2}= \left ( 5 \right )^{2}+\left ( PT \right )^{2}
\left ( PT \right )^{2}= 169-25
PT= {\sqrt{144}}= 12\, cm

PT and QT are tangents from same point
  \therefore PT = QT = 12 cm
AT = PT – PA 
AT = 12 – PA        ..…(i)
Similarly BT = 12 – QB          ..…(ii)
Since PA, PF and BF, BQ are tangents from point A and B respectively.
Hence, PA = AE              ..…(iii)
BQ = BE     ..…(iv)
AB is tangent at point E
  Hence OE \perp AB
\angle AET= 180^{\circ}-\angle AEO   \left ( \because \, \angle AEO= 90^{\circ} \right )
\angle AET= 180^{\circ}-90^{\circ}
\angle AET= 90^{\circ}
ET = OT – OF
ET = 13 – 5
ET = 8 cm
In \bigtriangleupAET, using Pythagoras theorem
\left ( H \right )^{2}= \left ( B \right )^{2}+\left ( P \right )^{2}
\left ( AT \right )^{2}= \left ( AE \right )^{2}+\left ( ET \right )^{2}
\left ( 12-PA \right )^{2}= \left ( PA \right )^{2}+\left ( 8 \right )^{2}  [using (i) and (ii)]
144+\left ( PA \right )^{2}-24\left ( PA \right )-\left ( PA \right )^{2}-64= 0
\left [ \because \left ( a-b \right )^{2}= a^{2}+b^{2}-2ab \right ]
80-42\left ( PA \right )= 0
24\left ( PA \right )= 80
\left ( PA \right )= \frac{80}{24}= \frac{10}{3}\, cm
\therefore \; AE= \frac{10}{3}\, cm\, \left [ using\left ( iii \right ) \right ]
Similiraly BE= \frac{10}{3}\, cm
AB= AE+BE= \frac{10}{3}+\frac{10}{3}= \frac{20}{3}\, cm

 

 

 

                        

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