In figure, 0 is the centre of a circle of radius 5 cm,

Given, OT = 13 cm and OP = 5 cm

Since, if we drawn a line from the centre to the tangent of the circle. It is always  perpendicular to the tangent i.e., OP⊥PT.

In right angled $\triangle O P T_{1} \quad O T^{2}=O P^{2}+P T^{2}$

[by Pythagoras theorem, (hypotenuse) $\left.^{2}=(\text { base })^{2}+(\text { perpendicular })^{2}\right]$

$\Rightarrow \quad P T^{2}=(13)^{2}-(5)^{2}=169-25=144$

$\Rightarrow \quad P T=12 \mathrm{~cm}$

Since the lenath of pair of tangents from an external point $T$ is equal.

$\therefore \quad Q T=12 \mathrm{~cm}$

Now. $\quad T A=P T-P A$

$\Rightarrow \quad T A=12-P A$ ..(i)

and $T B=Q T-Q B$

$\Rightarrow \quad T B=12-Q B \quad$...(ii)

Again, using the property, length of pair of tangents from an external point is equal.

$\therefore \quad P A=A E$ and $Q B=E B$ ....(iii)

$\therefore$$O T=13 \mathrm{~cm}$ 

$\Rightarrow \quad E T=13-5$

$\Rightarrow \quad E T=8 \mathrm{~cm}$

Since, $A B$ is a tangent and $O E$ is the radius.

$\Rightarrow$$O E \perp A B$

$\Rightarrow \quad \angle O E A=90^{\circ}$

$\therefore \quad \angle A E T=180^{\circ}-\angle O E A \quad$ [linear pair]

$\Rightarrow \quad \angle A E T=90^{\circ}$

Now, in right angled $\triangle A E T$,

$(A T)^{2}=(A E)^{2}+(E T)^{2}$ [by Pythagoras theorem]

$\Rightarrow \quad(P T-P A)^{2}=(A E)^{2}+(8)^{2}$

$\Rightarrow \quad(12-P A)^{2}=(P A)^{2}+(8)^{2}$ [from Eq. (iii)]

$\Rightarrow \quad 144+(P A)^{2}-24 \cdot P A=(P A)^{2}+64$

$\Rightarrow \quad P A=\frac{10}{3} \mathrm{~cm}$

 $\therefore$ $A E=\frac{10}{3} \mathrm{~cm}$ [from Eq. (iii)]

Join $O Q$.

Simitarly $B E=\frac{10}{3} \mathrm{~cm}$

Hence, $A B=A E+E B$

$=\frac{10}{3}+\frac{10}{3}$

$=\frac{20}{3} \mathrm{~cm}$

Hence,the required length $A B$ is $\frac{20}{3} \mathrm{~cm}$.