Question

In the given 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.

Answer 1

$OP=OQ=5cm$

$OT=13cm$

$OP$ and $PT$ are radius and tangent respectively at contact point P.

$\therefore \angle OPT={90}^{\circ }$

So, by pythagoras theorem, in right angled $\Delta OPT$,

$P{T}^2=O{T}^2-O{P}^2={13}^2-5^2=169-25=144$

$\Rightarrow PT=12cm$

$AP$ and $AE$ are two tangents from an external point A to a circle.

$\therefore AP=AE$

$AEB$ is tangent and $OE$ is radius at contact point E.

So, $AB\perp OT$ ___(i)

So, by Pythagoras theorem, in right angled. $\Delta AET$.

$A{E}^2=A{T}^2-E{T}^2$

$\Rightarrow A{E}^2=(PT-PA{)}^2-[TO-OE{]}^2$

$=(12-AE{)}^2-(13-5{)}^2$

$\Rightarrow A{E}^2=(12{)}^2+(AE{)}^2-2\left(12\right)\left(AE\right)-(8{)}^2$

$\Rightarrow A{E}^2-A{E}^2+24AE=144-64$

$\Rightarrow 24AE=80$

$\Rightarrow AE=\frac{80}{24}cm$

$\Rightarrow AE=\frac{10}{3}cm$

REF. IMAGE

In $\Delta TPO$ and $\Delta TQO$,

$OT=OT$ [common]

$PT=QT$ [Tangents from T]

$OP=OQ$ [Radii of same circle]

$\therefore \Delta TPO\cong \Delta TQO$ [By SSS criterion of congruence]

$\Rightarrow \angle 1=\angle 2$ ___(ii) [CPCT]

In $\Delta ETA$ and $\Delta ETB$,

$ET=ET$ [Common]

$\angle TEA=\angle TEB={90}^{\circ }$ [From (i)]

$\angle 1=\angle 2$ [CPCT] [From (ii)]

$\therefore \Delta ETA\cong \Delta ETB$ [By ASA criterion of congruence]

$\Rightarrow AE=BE$ [CPCT]

$\Rightarrow AB=2AE=2\times \frac{10}{3}$

$\Rightarrow AB=\frac{20}{3}cm$

Hence, the required length is $\frac{20}{3}cm$.