Perpendicular pair of circles

Geometry solution: If we have two unit circles, their tangents being perpendicular at intersection, then their normal vectors are also perpendicular at intersection. To find half the area, we can take the area of the 90º arc, to be $\frac{\pi}{4}$ , and subtract the area of the isosceles right triangle, $\frac{1}{2}$ . Since there are two of these areas, the full area between the circles is: $2\cdot \left(\frac{\pi}{4}-\frac{1}{2}\right) = \frac{\pi}{2}-1 \approx 0.570796$