Area of cut-out of three circles Let three circles at different centers with different radii be given.
They might intersect as shown in the picture.

How to derive the area of the set (blue) in terms of the centers and radii?
Because of multiple overlappings formulas like this seem not to
be useful.
 A: 
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:


*

*$\alpha$ and $\beta$, the angles of the green and pink triangles in the diagram; these are $\alpha = \arccos\left(\frac{f}{a}\right)$ and $\beta = \arccos\left(\frac{g}{a}\right)$.

*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.

*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.
A: 
First center the big circle center at the original with three circle's radius fixed. Think about several conditions, there needed the specification of the other two circles center.
So you have the position of three centers and their radius.
The cases that were boring not discuss here. Just consider the case in the graph. If you think clearly, the question you ask was essentially about the area of the little triangle in the middle.
(Also notice the segment connect any two circles will be orthogonal to the edge of the triangle)
Then you can find the intersection of the line with the edge, the left question was obvious.
