# 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.

• Is the rightmost red dot the center of the rightmost circle? – quasi Jun 19 at 18:18
• No. That's just part of the line that's been separated because of the overlapping blue line – yolo Jun 19 at 18:34
• Isn't $r$ supposed to be the length of the radius? – quasi Jun 19 at 18:36
• r is the length of the radius – yolo Jun 19 at 18:38
• Then $r$ is not correctly shown in the diagram. – quasi Jun 19 at 18:57

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.

• I understand you reaching that equation. I only wonder though how you reached my numerical value from it without solving for s. Or did you simply just miss out the equation. – yolo Jun 20 at 7:06
• Or did you solve by iteration? – yolo Jun 20 at 7:11
• I reached a quartic formula by rearranging your bottom equation. I went online to solve it and it didn't produce that value – yolo Jun 20 at 7:27
• Oh wait the equation went over a line. – yolo Jun 20 at 9:59
• @yolo: By the form of the equation, an exact symbolic solution is almost certainly unobtainable. To solve numerically, I used Maple's built-in fsolve function. – quasi Jun 20 at 17:21

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 $$s/r =2 (1-\cos \theta_1) \approx 1.19205$$