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.

Answers (1)

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

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 $\bigtriangleup$ AET, 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 $BE= \frac{10}{3}\, cm$
$AB= AE+BE= \frac{10}{3}+\frac{10}{3}= \frac{20}{3}\, cm$