Let AB be a chord of length 12 of the circle (x - 2) ^ 2 + (y + 1)...

Plotting the diagram of given value we have,

Now given, radius $= \dfrac{\sqrt{169}}{\sqrt{4}} = \dfrac{13}{2}$

Now by diagram $M$ is the midpoint of $AB$ so, $AM = 6$

Now in $∆ AMP$, $PM = 6cot\theta$

And in $∆ OAM$

$\sin \left(90 ° - \theta\right) = \dfrac{6}{\dfrac{13}{2}}$

$⇒ cos\theta = \dfrac{12}{13}$

$PM = 6 \times \dfrac{12}{5} = \dfrac{72}{5}$

$⇒ 5PM = 72$