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December 21

Calculate distance between centers of circles given area of intersections

I need help checking my work. I'm working on a representation of multiple regression effect sizes using a Venn Diagram for educational policy research. I see that the area of a geometric lens has a closed form solution. Given a Venn diagram made from circles $A$, $B$, and $C$ with centers $X$, $Y$, and $Z$, that $r=R=\sqrt{1/\pi }$, and $A_{A\cap B}=0.23$, then $d_{XY}\approx 0.429$. Can someone validate this? The Venn diagram is represented at this link here.Schyler (exquirito veritatem bonumque) 04:07, 21 December 2022 (UTC)

Circle $C$ has no role. I cannot replicate these numbers. When each circle passes through the centre of the other circle, so the circle centres are $r\approx 0.56419$ apart, a rhombus of two equilateral triangles fits within the lens. The area of this rhombus is $\frac{\sqrt{3}}{2\pi }\approx 0.27566,$ so when $d_{XY}=0.429<r$, $A_{A\cap B}>0.27566,$ contradicting $A_{A\cap B}\approx 0.23.$ My calculations give me that distance $d_{XY}=0.429$ gets you $A_{A\cap B}\approx 0.52785.$ Conversely, to get lens area $A_{A\cap B}=0.23,$ I find we need $d_{XY}\approx 0.73938.$  --Lambiam 07:38, 21 December 2022 (UTC)

Yes, circle $C$ has no role. Let $\theta$ be the angle at X (or Y) between the lines to the junctions of ABD and of CEFG. The lens has area $A=R^2(\theta -\sin \theta )$, where $R=\sqrt{1/\pi }$, and the diagram says $A_{A\cap B}$ (D+G) has area 0.03, so $\theta -\sin \theta =0.03\pi$. That gives $\theta \approx 0.8235$ (in radians). The length from X (or Y) to the midpoint of XY is $R\cos (\theta /2)$, and twice this gives $d_{XY}\approx 1.0341$. My trig is rusty and I may have made some errors, but I think the principle and the order of magnitude are right. If $A_{A\cap B}$ were 0.23, we'd get $\theta \approx 1.1377$ and $d_{XY}\approx 0.9507$. Certes (talk) 13:31, 21 December 2022 (UTC)

For $\theta =0.8235,$ we have $(\theta -\sin \theta )/\pi \approx 0.02864,$ not quite $0.03$ but at least in the ballpark. But in the second case, you missed a factor $\pi$: for $\theta =1.1377,$ we have $\theta -\sin \theta \approx 0.23$ while $(\theta -\sin \theta )/\pi \approx 0.07322.$  --Lambiam 15:52, 21 December 2022 (UTC)

Here are the calculations for lens area $0.23,$ step by step:

$\begin{array}{ccc}\theta & =& 1.71254\\ \sin \theta & =& \sin 1.71254& \approx & 0.98997\\ \theta -\sin \theta & \approx & 1.71254-0.98997& \approx & 0.72257\\ (\theta -\sin \theta )/\pi & \approx & 0.72257/3.14159& \approx & 0.23000\\ \cos (\theta /2)& \approx & \cos 0.85627& \approx & 0.65526\\ 2R\cos (\theta /2)& \approx & 2\times 0.56419\times 0.65526& \approx & 0.73938\end{array}$

--Lambiam 16:03, 21 December 2022 (UTC)

Oops. Thanks, that looks more credible. Certes (talk) 18:40, 21 December 2022 (UTC)

Okay, well this is interesting. Yes, in the example, circle C has no role here. Someone said "When each circle passes through the centre of the other circle," but I do not think that is probable. Here were my steps:

Circles $A$, $B$, and $C$ have centers $X$, $Y$, and $Z$ and radii $r_A=r_B=r_C=\sqrt{1/\pi }$. The centers form $\mathrm{△}XYZ$ and intersect such that $Are{a}_{A\cap B}=0.03$, $Are{a}_{B\cap C}=0.11$, $Are{a}_{A\cap C}=0.23$

$d_1=\overline{\mathrm{X}\mathrm{Y}};d_2=\overline{\mathrm{X}\mathrm{Z}};d_3=\overline{\mathrm{Y}\mathrm{Z}}$

$Are{a}_{A\cap C}=r_A^2{\cos }^{-1}(\frac{d_1^2+r_A^2-r_C^2}{2d_1{r}_A})+r_C^2{\cos }^{-1}(\frac{d_1^2+r_C^2-r_A^2}{2d_1{r}_C})-2\mathrm{\Delta }$

$\mathrm{\Delta }=\frac{1}{4}\sqrt{(-d_1+r_A+r_B)(d_1-r_A+r_B)(d_1+r_A-r_B)(d_1+r_A+r_B)}$

Simplifying:

$\cancel{0.23=2(\frac{1}{\pi }{\cos }^{-1}(\frac{d_1^2}{2d_1\sqrt{\frac{1}{\pi }}}))-\frac{1}{2}\sqrt{(-d_1+2\frac{1}{\pi })(d_1+2\frac{1}{\pi })}}$

oh I see this was my problem, I think... I distributed the 2 onto $\frac{1}{4}\mathrm{\Delta }$

$\cancel{0.23=2(\frac{1}{\pi }{\cos }^{-1}(\frac{d_1^2}{2d_1\sqrt{\frac{1}{\pi }}}))-\frac{1}{4}\sqrt{(-d_1+2\frac{1}{\pi })(d_1+2\frac{1}{\pi })}}$

Update: well, here i am checking my own work again. i found another error. i deleted some values of d within $\mathrm{\Delta }$. I should know better than to ask for help right away... i can do this... but the doubt is strong in this one

$\cancel{0.23=2(\frac{1}{\pi }{\cos }^{-1}(\frac{d_1^2}{2d_1\sqrt{\frac{1}{\pi }}}))-\frac{1}{4}\sqrt{(-d_1+2\frac{1}{\pi })(d_1+2\frac{1}{\pi })}}$

no that's wrong too, $r=\sqrt{\frac{1}{\pi }}$... i got it mixed up in $\mathrm{\Delta }$ again...

You can shift two equal-sized circles such that each passes through the other's centre; I used this special case merely to easily establish bounds that were violated by a purported solution.

The function $f(\theta )=(\theta -\sin \theta )/\pi$ is transcendental, and one cannot hope to solve the equation $f(\theta )=x$ by a combination of algebraic and trigonometric manipulations for rational values of $x,$ except when $x$ is an integer, in which case the equation is solved by $\theta =x\pi .$  --Lambiam 09:01, 22 December 2022 (UTC)