Question

Question 11
In figure , O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 and OT intersect the circle at E, if AB is the tangent to the circle at E, find the length of AB.

Answer 1

Given OT= 13 cm and OP = 5cm
If we draw a line from the centre to the tangent of the circle it is always perpendicular to the tangent i.e . OP $\perp$ PT.
In right angled $\Delta$ OPT,
$O{T}^2=O{P}^2+P{T}^2$
[ by Pythagoras theorem:
$(hypotenuse{)}^2=(base{)}^2+\left(perpendicular{)}^2\right]$
$\Rightarrow P{T}^2=(13{)}^2-(5{)}^2=169-25=144$
$\Rightarrow$ PT = 12 cm
Since, the length of pair of tangents from an external point T is equal.
$\therefore$ QT = 12 cm
Now. TA = PT - PA …………(i)
$\Rightarrow$ TA = 12 - PA
And TB = QT - QB ……..(ii)
$\Rightarrow$ TB = 12 - QB
Again, using the property, length of pair of tangents from an external point is equal.
$\therefore$ PA = AE and QB = EB (iii)
OT = 13 cm (Given)
$\therefore$ ET = OT - OE [ $\therefore$ OE = 5cm = radius]
$\Rightarrow$ ET = 13 - 5
$\Rightarrow$ ET = 8cm
Since AB is a tangent OE is the radius
$\therefore$ OE $\perp$ AB
$\Rightarrow$ OEA = ${90}^{\circ }$ [ linear pair]
$\angle AET+\angle OEA={180}^{\circ }$
$\therefore \angle AET={180}^{\circ }-\angle OEA$
$\Rightarrow \angle AET={90}^{\circ }$
Now, in right angled $\Delta$ AET,
$(AT{)}^2=(AE{)}^2+(ET{)}^2$ [by Pythagoras theorem]
$\Rightarrow (PT-PA{)}^2=(AE{)}^2+(8{)}^2\Rightarrow (12-PA{)}^2=(PA{)}^2+(8{)}^2\left(fromeq\right(iii\left)\right]\Rightarrow 144+(PA{)}^2-24PA=(PA{)}^2+64\Rightarrow 24PA=80$
$\Rightarrow PA=\frac{10}{3}cm\Rightarrow AE=\frac{10}{3}cm\left[from\right(iii\left)\right]BE=\frac{10}{3}cmAB=AE+EB=\frac{10}{3}+\frac{10}{3}=\frac{20}{3}cm$
Hence the required length AB is $\frac{20}{3}cm$