Question

$O$ is the centre of a circle of radius $5cm$. $T$ is a point such that $OT=13cm$ and $OT$ intersects the circle at $E$. If $AB$ is the tangent to the circle at $E$, find length of $AB$.

Answer 1

Clearly $\angle OPT={90}^o$

Applying Pythagoras in $△OPT$, we have

$O{T}^2=O{P}^2+P{T}^2$

$\Rightarrow {13}^2=5^2+P{T}^2$

$\Rightarrow P{T}^2=169-25=144$

$\Rightarrow PT=12cm$

Since lengths of tangents drawn from a point to a circle are equal. Therefore,

$AP=AE=x\left(say\right)$

$\Rightarrow AT=PT-AP=(12-x)cm$

Since $AB$ is the tangent to the circle$E$. Therefore, $OE\perp AB$

$\Rightarrow \angle OEA={90}^o$

$\Rightarrow \angle AET={90}^o$

$\Rightarrow A{T}^2=A{E}^2+E{T}^2$ [Applying Pythagoras Theorem in $△AET$]

$\Rightarrow (12-x{)}^2=x^2+(13-5{)}^2$

$\Rightarrow 144-24x+x^2=x^2+64$

$\Rightarrow 24x=80$

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

Similarly, $BE=\frac{10}{3}cm$

$\therefore AB=AE+BE=(\frac{10}{3}+\frac{10}{3})cm=\frac{20}{3}cm$