Ratio between the width of the intersection of two identical intersecting circles and radius, when the intersection is $\frac{\pi r^2}{2}$ Or more visually, if all sections of the below diagram were equal in area and the circles are identical, what is the ratio of s and r, or what is s in terms of r.

I came up with an equation using trigonometry and pythagoras. half the height height of the intersection is $\sqrt{r^2-\left(r-\frac{s}{2}\right)^2}$ where $r-\frac{s}{2}$ is the distance between a circle radius and the centre of the height of the intersection. From there I could work out the full height, then the area of the sector formed from the height as a chord and from that the area of the intersection, of which I know is $\frac{\pi r^2}. {2}$ due to the fact that all areas are equal. After working out the area of the triangle formed by the height and two radii, I found the angle of the sector with trig ($2\cos^{-1}\left(\frac{r-\frac{s}{2}}{r}\right)$). In conclusion the resultant equation is (with $r=x$ and $s=y$):
$\frac{\pi x^2}{2}=2\left(\frac{2\cos^{-1}\left(\frac{x-\frac{y}{2}}{x}\right)}{2\pi}\pi x^2-\frac{2\sqrt{x^2-\left(x-\frac{y}{2}\right)^2}\left(x-\frac{y}{2}\right)}{2}\right)$
I plugged this formula into desmos and recieved a straight line with a gradient close to $\frac{360000457}{302000000}$ but I wish to know the exact value.
 A: Without loss of generality, we can assume $r=1$.

For convenience, position the circles vertically, with centers on the $y$-axis at the points $(0,h)$ and $(0,-h)$.

Then the equation of the lower circle is
$$x^2+(y+h)^2=1$$
Solving for $y$, and noting that
$$h=1-\frac{s}{2}$$
we get that the upper half of the lower circle has the equation
$$y=-1+\frac{s}{2}+\sqrt{1-x^2}$$
Letting $(-a,0)$ and $(a,0)$ be the points where the circles interect, we get
$$a={\small{\frac{1}{2}}}\sqrt{s(4-s)}$$
hence the area of the region where the disks overlap is
\begin{align*}
&
4
\int_0^
{
{\Large{
\frac{1}{2}\sqrt{s(4-s)}
}}
}
\!
\left(
-1+\frac{s}{2}+\sqrt{1-x^2}
\right)
\;dx\\[6pt]
=\;&
2\sin^{-1}\left({\small{\frac{1}{2}}}\sqrt{s(4-s)}\right)-{\small{\frac{1}{2}}}(2-s)\sqrt{s(4-s)}\\[4pt]
\end{align*}
Solving the equation
$$
2\sin^{-1}\left({\small{\frac{1}{2}}}\sqrt{s(4-s)}\right)-{\small{\frac{1}{2}}}(2-s)\sqrt{s(4-s)}
=\pi/2$$
numerically yields 
$$s\approx 1.192054493$$
which matches the ratio you found.
A: Consider a half of the lens shaped lune cut along a central vertical line. Sector angle at center of either circle is $2 \theta$
$$ \frac{2 \theta}{2 \pi} .\pi r^2 -\frac12 r \sin \theta. r \cos\theta = \pi r^2/4$$
Simplifying 
$$ \theta - \sin \theta \cos \theta =\frac{\pi}{4} $$
A transcendental equation this is.. can be solved by numerical methods e.g., Regula_Falsi, Newton-Raphson etc. We get after iteration $$ \theta_1 \approx 1.15494  $$
Draw a horizontal line at mid of lens shape.
Its length is
$$ s/r =2 (1-\cos \theta_1) \approx 1.19205 $$
