in a given figure o is …

Clearly{tex}\angle O P T{/tex} = 90°
Applying Pythagoras in{tex}\Delta O P T{/tex}, we have
OT²=OP² +PT²
{tex}\Rightarrow {/tex}13² = 5² + PT²
{tex}\Rightarrow {/tex} PT² = 169 - 25 = 144
{tex}\Rightarrow {/tex} PT = 12 cm
Since lengths of tangents drawn from a point to a circle are equal. Therefore,
AP = AE = x(say)
{tex}\Rightarrow {/tex}  AT = PT - AP = (12 - x)cm
Since AB is the tangent to the circle E. Therefore, {tex}O E \perp A B{/tex}
{tex}\Rightarrow \quad \angle O E A = 90 ^ { \circ }{/tex}
{tex}\Rightarrow \quad \angle A E T = 90 ^ { \circ }{/tex}[Applyng Pythagoras Theorem in {tex}\Delta A E T{/tex}]
{tex}\Rightarrow {/tex} (12 - x)² = x² + (13 - 5)²
{tex}\Rightarrow {/tex} 144 - 24x + x² = x² + 64
{tex}\Rightarrow {/tex} 24x = 80
{tex}\Rightarrow \quad x = \frac { 10 } { 3 } \mathrm { Cm }{/tex}
Similarly, BE {tex}= \frac { 10 } { 3 } \mathrm { cm }{/tex}
{tex}\therefore{/tex} AB = AE + BE = {tex}\left( \frac { 10 } { 3 } + \frac { 10 } { 3 } \right) \mathrm { cm }{/tex}
{tex}= \frac { 20 } { 3 } \mathrm { cm }{/tex}