Two circles of radius 13 cm and 15 cm intersect at A and B. What is the distance between the centres of the circles if AB = 24 cm?

Given: AB = 24 cm

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

Since the line joining the centres of the circles bisects the common chord.

$\therefore \mathrm{AC}=\mathrm{BC}=\frac{24}{2}=12 \mathrm{cm}$

In $∆\mathrm{AOC}, \angle \mathrm{OCA}= {90}^{\mathrm{o}}$

By Pythagoras theorem, 

${\mathrm{OA}}^2={\mathrm{OC}}^2+{\mathrm{AC}}^2$

$\Rightarrow {15}^2={\mathrm{OC}}^2+{12}^2$

$\Rightarrow {\mathrm{OC}}^2=225-144$

$\Rightarrow {\mathrm{OC}}^2=81$

$\Rightarrow \mathrm{OC}=9 \mathrm{cm}$

Similarly, in $∆\mathrm{AO}\text{'}\mathrm{C}, \angle \mathrm{O}\text{'}\mathrm{CA}={90}^{\mathrm{o}}$

By Pythagoras theorem, 

$\mathrm{O}\text{'}{\mathrm{A}}^2=\mathrm{O}\text{'}{\mathrm{C}}^2+{\mathrm{AC}}^2$

$\Rightarrow {13}^2=\mathrm{O}\text{'}{\mathrm{C}}^2+{12}^2$

$\Rightarrow \mathrm{O}\text{'}{\mathrm{C}}^2=169-144$

$\Rightarrow \mathrm{O}\text{'}{\mathrm{C}}^2=25$

$\Rightarrow \mathrm{O}\text{'}\mathrm{C}=5 \mathrm{cm}$

Now,  $\mathrm{OO}\text{'}=\mathrm{OC}+\mathrm{O}\text{'}\mathrm{C}$

$\Rightarrow \mathrm{OO}\text{'}=9+5=14 \mathrm{cm}$

Hence, the distance between the centres of the circles is 14 cm.