Wikipedia:Reference desk/Archives/Mathematics/2022 December 21

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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 ${\tfrac {\sqrt {3}}{2\pi }}\approx 0.27566,$ 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 $A=R^{2}\left(\theta -\sin \theta \right)$, 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:

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

--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 $\triangle XYZ$ and intersect such that $Area_{A\cap B}=0.03$, $Area_{B\cap C}=0.11$, $Area_{A\cap C}=0.23$

$d_{1}={\overline {\rm {XY}}};d_{2}={\overline {\rm {XZ}}};d_{3}={\overline {\rm {YZ}}}$

$Area_{A\cap C}=r_{A}^{2}\cos ^{-1}({\frac {d_{1}^{2}+r_{A}^{2}-r_{C}^{2}}{2d_{1}r_{A}}})+r_{C}^{2}\cos ^{-1}({\frac {d_{1}^{2}+r_{C}^{2}-r_{A}^{2}}{2d_{1}r_{C}}})-2\Delta$

$\Delta ={\frac {1}{4}}{\sqrt {(-d_{1}+r_{A}+r_{B})(d_{1}-r_{A}+r_{B})(d_{1}+r_{A}-r_{B})(d_{1}+r_{A}+r_{B})}}$

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 ${\frac {1}{4}}\Delta$

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

Update: well, here i am checking my own work again. i found another error. i deleted some values of d within $\Delta$. 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 $\Delta$ 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)