Two tangent circles inscribed in a rectangle (Compute the area) Consider two circles with a diameter equal to $a$, externally tangent to each other, and whose centers are at the same height. Those circles are inscribed inside a rectangle of length $2a$ and height $a$. This is a sketch I made for this problem (please, forgive my unsteady handwriting):

I am asked to calculate the shaded area.
I can do it using:


*

*Symmetry. This is the easiest way in my opinion, since we know the area of each circle ($\pi a^2/4$) and the area of the rectangle ($2a^2$), which gives us: $\boxed{A_{\text{shaded}}=\dfrac{4-\pi}{4} a^2}$

*Mathematical functions. We can set the origin at the bottom left corner, calculate each of the circles' analytical functions (as well as that of the line), compute the intersections, and make use of definite integrals to compute the final area. It will yield the exact same result as the above, yet the process to achieve it would be much longer.


However, I'm not interested in any of these 2 methods (as they look pretty easy). I'm interested in finding a pure geometrical way to solve it. No functions. No symmetry as I used above. But pure geometric relationships.
I thought about drawing some lines from the center of each circle to each intersection, like this:

This would give us 4 areas hopefully easy to solve for ($S_1$, $S_1^{\prime\prime}$ and $S_2$, $S_2^{\prime\prime}$). Of course, $S_1=S_1^{\prime\prime}$ and $S_2=S_2^{\prime\prime}$ but as I said, I don't want to make use of symmetry (I'm a bit masochistic after all). So I thought about this solving scheme:


*

*$A_{\text{shaded}}=\frac12A_{\text{rectangle}}-S_2-(A_{\text{circle}}-S_2^{\prime\prime})$

*$A_{\text{sector}}=S_1+S_2$

*Since $S_1$ is a triangle, we could use some trigonometric relations in it to solve for $S_1$, and since the area of a circular sector is known, we could solve for $S_2$ (and $S_2^{\prime\prime}$, and very innocently find that actually $S_2=S_2^{\prime\prime}$)


The main problem I have is that I don't know the inner angle of the circle sectors, because of the (seemingly randomly placed) intersection points near the borders of the rectangle.
IF ONLY I were able to locate those intersection points using pure trigonometry and geometric relationships/theorems, the problem could be solved.
Any hints or ideas?
PD: Please, don't question my (foolish) decision of not making use of symmetry. I want to take this problem as a personal challenge. Obviously, if this were to be solved quickly I wouldn't scratch my head too much and go for the easy solution.
 A: From the lower left-hand corner of the rectangle (let's call that point $A$), consider the diagonal line as a secant line of the left-hand circle,
intersecting the circle at points $B$ and $C$, where $B$ is between $A$ and $C$.
Also consider one of the edges of the rectangle adjacent to $A$
as a tangent line touching the circle at $D$.
Then by a theorem about tangent and secant lines from a point outside a circle, we have the following relationship of the lengths of the segments
from $A$ to each of the three points $B$, $C$, and $D$:
$$
(AD)^2 = AB \times AC. \tag1
$$
It is easy to find that $AD = \frac12a$.
Now let $E$ be the midpoint of the bottom side of the rectangle;
then $AC$ is the hypotenuse of right triangle $\triangle AEC$,
which has legs $a$ and $\frac12a$, and therefore
$AC = (\frac12\sqrt5)a$.
We can then use Equation $(1)$ to find the length $AB$,
so we can find the length of the chord $BC$; from that chord and
the radius of the circle we can get the angle of $S_1$ at the
center of the circle.
A: I don't know if this is what you want but.
The triangle marked $S_1$ is a isosceles triangle with base angle  $= \tan^{-1} \frac 12$,
This makes the vertex angle (v) equal to  $v = \pi - 2\tan^{-1} \frac 12$
Which means that $A_{sector} = S_1 + S_2 = \frac 12(\frac a2)^2 (\pi - 2\tan^{-1} \frac 12)$
$A_{S_1} = $$(\frac a2)^2 \frac 12 \sin v\\
 (\frac a2)^2 \sin (2\tan^{-1} \frac 12)\\
(\frac a2)^2 2 \frac 1{\sqrt5} \frac 2{\sqrt5}\\
a^2 \frac 15$
A: Is analytic geometry all right?
Take the center of the rectangle as the origin.  The two circles have the equations
$$(x \pm (a/2))^2 + y^2 = a^2/4.$$
The diagonal has the equation
$$2x=y.$$
Solving for these will give you the four points of intersection.
To get the central angles then you can use the Law of Sines after finding the length of the chords.
A: I think you can consider the full rectangle area 2a^2 subtracting the areas of two circles then devide the result by 2.
A: Jose Perez
You can find the angle between diagonal and length by using
x=inv.tan (2R/4R) 
It is equal to the angle of triangle s1 (alternating angles). And there you go, angle of sector s 180-2x. 
