Area of cut-out of three circles

First some names: the circle we want to get the area of a portion of is centered at $A$, with radius $a$; we also want the other two circles, centered at $B$ and $C$ and with radii $b$ and $c$ respectively.

The first task is to find the two radical lines, the first between $A$ and $B$, and the second between $A$ and $C$; on the way we'll pick up a few other things.

Find $d = |B-A|$, the distance between $B$ and $A$. Similarly find $e = |C-A|$.

Then, we find the distance from $A$ to the $A$-$B$ radical line: $f = \frac{a^2-b^2+d^2}{2d}$ and similarly $g$ using $c$ and $e$.

With $f$ and $g$ we can find several things:

1. $\alpha$ and $\beta$, the angles of the green and pink triangles in the diagram; these are $\alpha = \arccos\left(\frac{f}{a}\right)$ and $\beta = \arccos\left(\frac{g}{a}\right)$.
2. The areas of the green and pink triangles: $\color{green}{\frac{f\sqrt{a^2 - f^2}}{2}}$ and $\color{magenta}{\frac{g\sqrt{a^2-g^2}}{2}}$ respectively.
3. the radical lines, which intersect in the radical center. This bit's a little more involved.

We'll need $h$ and $i$, unit vectors in the directions of the two center lines: $h = \frac{B-A}{|B-A|}$ and $i = \frac{C-A}{|C-A|}$.

While we're here, $\gamma = \arccos(h\cdot i)$ is the angle between the two centerlines, and $\theta = 2\pi - \alpha - \beta - \gamma$ is the angle of the remaining circular sector, so $\color{red}{\frac{a^2\theta}{2}}$ is the area of that sector.

Our lines are $h\cdot v = h\cdot A + f$ and $i\cdot v = i\cdot A + g$; we can then solve for $v$ using standard line intersection work.

Now we merely need the distance from $v$ to $A$ $j = |v-A|$ and we can find the areas of blue and yellow: $\color{blue}{\frac{f\sqrt{j^2 - f^2}}{2}}$ and $\color{goldenrod}{\frac{g\sqrt{j^2-g^2}}{2}}$.

Add it all up and we get

$$\frac{\color{red}{a^2\theta} + \color{green}{f\sqrt{a^2 - f^2}} + \color{blue}{f\sqrt{j^2 - f^2}} + \color{goldenrod}{g\sqrt{j^2-g^2}} + \color{magenta}{g\sqrt{a^2-g^2}}}{2}$$

Which might stand a little more simplifying I guess.

Warning: There are layouts where the blue section cuts into the green, or the yellow into the magenta, and in these cases you'll have to subtract. In addition it's possible for $v$ to be outside all three circles; I'm not sure what you'd actually want in that case.