It's a straightforward but messy exercise to find the area of the "lens" formed by the intersection of the two circles. If the radii are the same, consider a circle with center $(0,0)$ and radius 1 and another with center $(d,0)$ and radius $1$. Then the area of the lens divided by $\pi$ is the percentage you are looking for. The circles are
$$C_0: \ \ x^2+y^2 = 1,\ \ \ C_d: \ \ (x-d)^2 +y^2 = 1.$$
Solving for the intersection points we find
$$ P_1 = (d/2,\sqrt{4 - d^2}/2 , \ \ P_2 = (d/2,-\sqrt{4 - d^2}/2) $$
The segment $P_1 P_2$ is a chord that cuts both circles. You can find formulas for the area between each circle and the chord -- then add them to obtain the area of the lens.
The area between the chord and $C_0$ is
$$A_0 = \arccos(d/2)- d\sqrt{4 - d^2}/4.$$
The area between the chord and $C_d$ is identical by symmetry
$$A_d = \arccos(d/2)- d\sqrt{4 - d^2}/4.$$
Hence the percentage of overlap is
$$O = \frac{2 \arccos(d/2)- d\sqrt{1 - d^2/4}}{\pi}.$$
In order to find $d$ for a specified value of $O$, you have to solve a non-linear equation numerically. There is no closed form solution.
If $O = 0.5$ then $d = 0.8079455...$ approximately.
Let's start by finding the combined area of the two larger circles. The area of one of the larger circles, which I will call $A_O$, is:
$$
\begin{align*}
A_O &= \frac14\pi d^2\\
&= \frac14\pi(4.25)^2\\
&\approx14.186
\end{align*}
$$
Thus the area of both circles, ignoring their intersection for a moment, is $2A_O$. To find the area between them, we can use the formula for the area of a symmetric lens,
$$A_\text{lens} = R^2(\theta-\sin\theta), \tag{*}\label{*}$$
where $R$ is the radius of the circles forming the lens, and $\theta$ the central angle in radians. We know $R = 4.25$, but we need to find $\theta$. To do so, draw in a few lines on the diagram:
(Note that I left out the smaller circle for simplicity's sake.) In the diagram, $\theta = m\angle AOB = 2\alpha$, and $R = OA = OC = OB$. We can find $\theta$ through $\triangle AOD$. The length of line segment $OD$ is equal to $OC-CD$, or the radius minus half the distance between the two circles. We also know the length of line segment $OA$, since it is the radius, so we can write:
$$
\begin{align*}
\alpha &= \arccos\left(\frac{OD}{OA}\right)\\
&= \arccos\left( \frac{2.125-\frac12(1.34)}{2.125} \right)\\
&\approx .816596 \ \text{radians}
\end{align*}
$$
Since $\theta = 2\alpha$, $\theta=1.63319$. We can now use $\eqref{*}$:
$$
\begin{align*}
A_\text{lens} &= R^2(\theta-\sin\theta) \\
&= (2.125)^2(1.63319-\sin1.63319)\\
&\approx 2.868
\end{align*}
$$
Thus, the area between the two larger circles, $A_L$, is:
$$
\begin{align*}
A_L &= 2A_O-A_\text{lens}\\
&= 2(14.186)-2.868\\
&= 25.504
\end{align*}
$$
Now we have to deal with the two small regions created by the smaller circle. For these, I will be using the (rather lengthy) formula for the area of an asymmetric lens,
$$
\begin{align*}
A_\text{asy} = & \ r^2\arccos\left( \frac{d^2+r^2-R^2}{2dr} \right)
+ R^2\arccos\left( \frac{d^2+R^2-r^2}{2dR} \right) - \\
& \frac12 \sqrt{(-d+r+R)(d+r-R)(d-r+R)(d+r+R)}, \tag{**}\label{**}
\end{align*}
$$
where $r$ is the radius of the smaller circle, $R$ is the radius of the larger circle, and $d$ is the distance between the centers of the two circles. Here, $r=1.64$, $R = 2.125$, and $d=1.455$. Observe the diagram below.
The light blue region is the asymmetric lens we will be finding the area of, and we are looking for the area of the two green regions. Using \eqref{**}, we find the area of one asymmetric lens to be $A_\text{asy}\approx5.650$, and thus the area of two asymmetric lenses is $2A_\text{asy}=11.300$.
Notice that the intersection of the two asymmetric lenses in the diagram is the symmetric lens from before. Thus, the area of the light blue and dark blue region is:
$$
\begin{align*}
A_\text{blue} &= 2A_\text{asy}-A_\text{lens}\\
&=11.300-2.868\\
&=8.432
\end{align*}
$$
Now, the area of the green regions is equal to the area of the small circle minus the area of the blue regions. The area of the small circle is:
$$
\begin{align*}
A_o &= \pi r^2 \\
&= \pi(1.64)^2\\
&\approx 8.450
\end{align*}
$$
And so the area of the green regions is:
$$
\begin{align*}
A_\text{green} &= A_o-A_\text{blue}\\
&=8.450-8.432\\
&=0.018
\end{align*}
$$
Finally, the area of the entire shape is the area of the two large circles plus the area of the green regions, which is:
$$
\begin{align*}
A &= A_L + A_\text{green} \\
&= 25.504 + 0.018\\
&= \boxed{25.522 \text{ mm}^2}
\end{align*}
$$
Best Answer
The lenticular intersection of the two circles consists of two circular segments, the areas of which are given by the reasonably well-known formula $\frac12 R^2(\theta+\sin\theta)$, where $\theta$ is the angle of arc covered by the segment. The appearance of both an angle and a trig function of that angle in that expression doesn’t bode well for finding a nice closed-form solution.
Referring to the above illustration, the intersection area is equal to the sum of the areas of the two circular sectors less the area of the kite formed by the radii: $$\frac12 r_1\theta_1 + \frac12 r_2\theta_2 - hd. \tag 1$$ The altitude $h$ and the proportion $\lambda$ of the distance between the centers the the line of intersection are easily found via the Pythagorean theorem: $$r_1^2-\lambda^2d^2 = h^2 = r_2^2 - (1-\lambda)^2d^2$$ from which $$\lambda = {d^2+r_1^2-r_2^2 \over 2d^2} \tag 2$$ and $$1-\lambda = {d^2-r_1^2+r_2^2 \over 2d^2}. \tag 3$$ We also have $\cos{\frac{\theta_1}2} = {\lambda d\over r_1}$ and $\sin{\frac{\theta_1}2} = \frac h{r_1}$, and similarly for $\theta_2$. Solving for the two angles and substituting everything back into expression (1) produces a rather ugly expression for the intersection area in terms of $r_1$, $r_2$ and $d$. It involves inverse trig functions of functions of $d$ and the square root of a rational expression in $d$, so there’s little chance of a closed-form solution, however complex.
I think the best bet for your code is to compute numerical approximations to $d$ and $\lambda$. $d$ is bounded above by $r_1+r_2$ and below by $|r_1-r_2|$, so a simple binary search might work pretty well. You could make a first guess at $d$ by approximating the intersection area with the area of the inscribed kite and solving the resulting equation that doesn’t involve any inverse trig functions, but that solution is likely to be fairly complex, too, and I doubt that it would buy you much, if anything, in terms of computational efficiency.