Overlapping Area of Two Circles on Diagonal

Square side = 2. Radii $r = 2/3$. Centers are on diagonal, touching sides $\implies$ distance from corner to center $C_1 = r\sqrt{2} = \frac{2\sqrt{2}}{3}$. Distance from opposite corner to $C_2 = \frac{2\sqrt{2}}{3}$. Total diagonal = $2\sqrt{2}$. Distance $C_1 C_2 = 2\sqrt{2} - 2 \times \frac{2\sqrt{2}}{3} = \frac{2\sqrt{2}}{3}$. Note $C_1 C_2 = r\sqrt{2} = \frac{2\sqrt{2}}{3}$. Intersection of two congruent circles of radius $r$ separated by distance $d = r\sqrt{2}$: The line connecting intersection points $P, Q$ forms a right triangle with $C_1, C_2$ because $C_1 P^2 + C_2 P^2 = r^2 + r^2 = 2r^2 = (r\sqrt{2})^2 = (C_1 C_2)^2$. So angle subtended at center by overlapping arc is $90^\circ$. Area of segment of one circle = Sector area - Triangle area = $\frac{\pi}{4} r^2 - \frac{1}{2} r^2 = r^2 (\frac{\pi - 2}{4})$. Total overlapping area = $2 \times \text{segment area} = 2 \times \frac{r^2(\pi-2)}{4} = \frac{r^2(\pi-2)}{2}$. Since $r = 2/3$, $r^2 = 4/9$. Overlapping area = $\frac{4/9(\pi - 2)}{2} = \frac{2(\pi - 2)}{9}$.