Given figure shows two circles passing through each other's centres. Find the area of the region enclosed by the two circles in terms of the common radius $r$.

Let the centres of the two congruent circles be A and B, each having radius $r$.

Since each circle passes through the other's centre, $AB = r$.

The common region is made of two identical circular segments.

For central angle consider $\triangle ACB$.

$AC = r, BC = r, AB = r$

Hence $\triangle ACB$ is equilateral.

Therefore, $\angle CAB = 60^\circ$

Similarly, $\angle CBA = 60^\circ$

So, each sector forming the overlap has angle $120^\circ$.

(because the segment is bounded by the major side inside the overlap).

For area of one sector,

$$\text{Area of a } 120^\circ \text{ sector: } A = \frac{\theta}{360^\circ} \pi r^2$$

With $\theta = 120^\circ$,

$$\text{sector area} = \frac{120^\circ}{360^\circ} \pi r^2 = \frac{1}{3} \pi r^2$$

Now, triangle ACB is equilateral with side $r$.

$$\text{Its area is } \frac{\sqrt{3}}{4} r^2.$$

$$\text{And segment area} = \frac{1}{3} \pi r^2 - \frac{\sqrt{3}}{4} r^2$$

There are two identical segments.

Hence common area:

$$2 \left( \frac{1}{3} \pi r^2 - \frac{\sqrt{3}}{4} r^2 \right) = \frac{2 \pi r^2}{3} - \frac{\sqrt{3}}{2} r^2$$