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I am looking for the area of the white region interior to a set of circles with radius A, oriented on the edge of a larger circle with radius B, spaced apart from each other with distance C. You can assume that C is less than 2 times A, so that each smaller circle overlaps with its neighbors. To clarify, I am looking for the area of the circle with radius b, not covered by small circles.

  • $\begingroup$ I pasted your figure in. The filename indicated you were after the white area in the middle but your description seems to indicate the total gray area. I will restore the white area, but please check and clarify. You can click the edit button at the bottom of the question to make it how you want. $\endgroup$ – Ross Millikan Apr 7 '18 at 14:50
  • $\begingroup$ Can we assume any minimum for $C$, say $C \ge A$? $\endgroup$ – Jens Apr 7 '18 at 15:24
  • $\begingroup$ Yes, c can only be >= a $\endgroup$ – Selav Apr 7 '18 at 15:54
  • $\begingroup$ Note that $B$ determines $C$ up to a discrete set of choices: $C=2B\sin\frac\pi n$. $\endgroup$ – Henning Makholm Apr 7 '18 at 16:09
  • $\begingroup$ Here is a quick demo that has g.kov's solution implemented along with a numerical solution demonstrated graphically: compassionate-allen-221713.bitballoon.com $\endgroup$ – Selav Apr 8 '18 at 3:22

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It's more convenient to use the number of circles $n$ instead of the distance between their centers.

Then \begin{align} \phi&=\tfrac\pi{n} , \end{align} the distance between the centers $|C_iC_{i+1}|=2\,R\sin\tfrac\phi2$ and the total area $S$ of the interior region consists of $n$ areas $S_p$ of petals $P_1OP_2$.

\begin{align} S&=n\,S_p ,\\ S_p &=2\,S_{\triangle OC_0P_1}-S_{\mathrm{seg}\,C_0P_1P_2} \\ &=|OC_0|\cdot|P_1F|-\tfrac\theta2\,|C_0P_1|^2 \\ &=R\cdot r_a\sin\tfrac\phi2-\tfrac\theta2\,r^2 \tag{1}\label{1} . \end{align}

In $\triangle P_1OF$, $\triangle P_1FC_0$, $\triangle P_1C_0G$

\begin{align} \angle C_0GO&=\tfrac\pi2 ,\\ \angle GOC_0&=\tfrac\phi2=\tfrac\pi{n} ,\\ \angle FP_1O&=\tfrac\pi2-\tfrac\phi2 ,\\ \angle FC_0P_1&=\tfrac\theta2 ,\\ \angle C_0P_1F&=\tfrac\pi2-\tfrac\theta2 ,\\ \angle GP_1C_0& =\tfrac\phi2+\tfrac\theta2 =\tfrac\pi{n}+\tfrac\theta2 ,\\ |C_0G|&=R\sin\tfrac\phi2=R\sin\tfrac\pi{n} , \end{align}

\begin{align} \sin\angle GP_1C_0&= \sin(\tfrac\pi{n}+\tfrac\theta2) =\tfrac{|C_0G|}{r} =\tfrac{R}r\,\sin\tfrac\pi{n} ,\\ \tfrac\theta2&= \arcsin\left(\tfrac{R}r\,\sin\tfrac\pi{n}\right)-\tfrac\pi{n} ,\\ \sin\tfrac\theta2&= \tfrac{R}{r}\,\sin\tfrac\pi{n}\, \left( \cos\tfrac\pi{n}-\sqrt{\tfrac{r^2}{R^2}-\sin^2\tfrac\pi{n}} \right) ,\\ r_a&=\frac{r}{\sin\tfrac\pi{n}}\,\sin\tfrac\theta2 \\ &= R\, \left( \cos\tfrac\pi{n}-\sqrt{\tfrac{r^2}{R^2}-\sin^2\tfrac\pi{n}} \right) . \end{align}


\begin{align} S(n,R,r)&= \pi\,r^2+ n\cdot \left( R^2\sin(\tfrac\pi{n})\, \left( \cos(\tfrac\pi{n})-\sqrt{\tfrac{r^2}{R^2}-\sin^2(\tfrac\pi{n})} \right) -r^2\, \arcsin\left(\tfrac{R}r\,\sin(\tfrac\pi{n})\right) \right) . \end{align}

Edit An example with (more-or-less) nice expression for the area.

Let $n=6$, $R=\sqrt3$, $r=1$. Then

\begin{align} r_a&=\sqrt3\left(\tfrac{\sqrt3}2-\sqrt{\tfrac13-\tfrac14} \right) =\tfrac32-\sqrt{1-\tfrac34}=1 ,\\ \tfrac\theta2&=\arcsin(\tfrac{\sqrt3}2)-\tfrac\pi6=\tfrac\pi6 ,\\ S(6,\sqrt3,1) &= \pi+6\cdot \left( \tfrac32\, \left( \tfrac{\sqrt3}2 -\sqrt{\tfrac{1}{3}-\tfrac{1}{4}} \right) -1\cdot \arcsin\tfrac{\sqrt3}2 \right) \\ &= \pi+ 9\, \left( \tfrac{\sqrt3}2 -\tfrac{\sqrt3}6 \right) -6\cdot \tfrac\pi3 \\ &= 3\sqrt3-\pi\approx 2.05 . \end{align}

In the image below four grid cells represent one square unit:

enter image description here

  • $\begingroup$ Great, neat answer. Thank you! $\endgroup$ – Selav Apr 8 '18 at 3:00

The area within the intersection of two circles is called a lens; when the circles each have radius $a$ and their centres are distance $d$ apart, the area of the lens is

$$L = a^2\pi-2a^2\arctan\left(\frac{d}{\sqrt{4a^2-d^2}}\right)-\frac{d}{2}\sqrt{4a^2-d^2}$$

Suppose there are $n$ small circles. Then the total area is the area of the circles minus the area of the lenses, i.e.:

$$n(\pi a^2 - L)$$

  • $\begingroup$ OP wanted the white area, so it would be $\pi B^2-nL$ except that the area common to two small circles and the large one is subtracted twice. If $C$ is small enough you can even get four way overlaps. $\endgroup$ – Ross Millikan Apr 7 '18 at 14:37
  • $\begingroup$ @RossMillikan Ah, I didn't think about four-way overlaps; good point. I thought at first that OP was looking for the white area, but on second reading interpreted the "region interior to the circles" as being within the small circles themselves, not interior to their outline. $\endgroup$ – Théophile Apr 7 '18 at 14:46
  • $\begingroup$ I pasted the graphic in. The title (which disappeared) was the area of the white region. I'll put that back. I believe that is what OP wanted. Sorry for the confusion. $\endgroup$ – Ross Millikan Apr 7 '18 at 14:49
  • $\begingroup$ @RossMillikan Okay, thanks for the clarification. Maybe it is easier to use calculus here? $\endgroup$ – Théophile Apr 7 '18 at 16:09

Below is an example figure with $n=8$ small circles

enter image description here

Given $n$ small circles, the white area can be determined as the area of the large circle with radius $B$ minus the areas of the $n$ green sectors with angle $\alpha_1$ and the $n$ blue sectors with angle $\alpha_2$, i.e.: $$\text {Area} = \pi B^2- nA^2\frac{(\alpha_1 + \alpha_2)}{2}$$ The problem now is to find $\alpha_1$ and $\alpha_2$. Finding $\alpha_1$ is easy using the Cosine Rule, as we have an isosceles triangle with sides $A$ and base $C$, which gives $$\cos \alpha_1 = 1- \frac{C^2}{2A^2}$$

Finding $\alpha_2$ is harder. But we can use the triangle with the angle $\alpha_3$ shown above. We know that $2\alpha_3 = \frac{2\pi}{n}$ and hence $\alpha_3 = \frac{\pi}{n}$. We also know two sides of the triangle and can therefore use the Sine Rule to find $\alpha_4$: $$\frac{\sin \alpha_3}{A}=\frac{\sin \alpha_4}{B}$$ which gives $$\sin \alpha_4= \frac{B}{A} \sin \alpha_3$$ Hence $$\alpha_4 =\pi-\text {arcsin} (\frac{B}{A} \sin \alpha_3)$$ and therefore $$\alpha_2 = 2(\pi-\alpha_3-\alpha_4)$$

I suspect there is a simpler answer to this question, but I couldn't immediately find it.


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