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I am trying to determine how to calculate the required radius of the smaller circles so they touch each other around the larger circle. (red box)

I would like to be able to adjust the number of smaller circles and the radius of the larger circle.

As an example:

$$\begin{align} R&=1.5\\ n&=9\\ r&=\,? \end{align}$$

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  • $\begingroup$ What do you mean with "around the big circle", exactly? Where are the centers of the small circles? $\endgroup$
    – ajotatxe
    Dec 16, 2014 at 20:19
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    $\begingroup$ Are you saying the centers of the small circles must lie on the circumference of the large one? $\endgroup$ Dec 16, 2014 at 20:23
  • $\begingroup$ !example 2 $\endgroup$
    – James
    Dec 16, 2014 at 20:26
  • $\begingroup$ Do two subsequent small circles have to intersect on the big circle or they just have to be tangent? $\endgroup$
    – Surb
    Dec 16, 2014 at 20:36
  • $\begingroup$ I am trying to get the smaller circles uniform around the bigger circle. Tangent would work. $\endgroup$
    – James
    Dec 16, 2014 at 20:42

3 Answers 3

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If you draw $n$ lines from the origin touching the small circles and $n$ lines from the origin to the center of each small circle you basically divide the $2 \pi$ angle into $2n$ equal angles, say $\theta$. Hence $\theta={\pi}/{n}$. Now a triangle with vertices the origin, the center of one of the small circles and a tangent point of the same circle is a right triangle since the tangent is perpendicular to the radius at the point of contact. You then have

$$ \sin \theta=\frac{r}{R} $$

Putting it all together

$$ r=R \sin \frac{\pi}{n} $$

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Another approach: Lets say we have $n$ small circles. Then the center points of the small circles form a regular $n$-gon, where the side length is $2r$. The radius of the big circle is the circumradius of the $n$-gon which is $R = \frac{2r}{2\sin(\pi/n)}$ so $r = R \sin(\pi/n)$

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  • $\begingroup$ OK but I was faster... $\endgroup$
    – Georgy
    Dec 19, 2014 at 23:07
  • $\begingroup$ Timestamp says the contrary, you were 3 min too late=) $\endgroup$
    – flawr
    Dec 19, 2014 at 23:34
  • $\begingroup$ Compare "edited Dec 16 @ 20:50" with "answered Dec 16 @ 20:48". Oh never mind, I'm just teasing you! $\endgroup$
    – Georgy
    Dec 19, 2014 at 23:44
  • $\begingroup$ So what=) At least we agree that this is a rather nice way of solving it! $\endgroup$
    – flawr
    Dec 19, 2014 at 23:48
  • $\begingroup$ That we do sir :) $\endgroup$
    – Georgy
    Dec 19, 2014 at 23:52
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Suppose that the center $c_k,k=1,\ldots,n$ of the small circles are placed equidistantly on the bigger circle.

Then we have $c_k=\left(R\sin(2\pi\frac{k}{n}),R\cos(2\pi\frac{k}{n})\right), k = 1,\ldots,n$. So for $k=1,\ldots,n$ we must have $r=\frac{\|c_{k}-c_{k+1}\|_2}{2}$, since two circles with same radius are tangent if their radius is the half of the distance between their centers. In particular $$r=\frac{\|c_{n-1}-c_{n}\|_2}{2}=\frac{1}{2}\sqrt{\big(R\sin(2\pi)-R\sin\big(2\pi\frac{n-1}{n}\big)\big)^2+\big(R\cos(2\pi)-R\cos\big(2\pi\frac{n-1}{n}\big)\big)^2} \\ =\frac{R}{2}\sqrt{\sin\big(2\pi\frac{n-1}{n}\big)^2+\big(1-\cos\big(2\pi\frac{n-1}{n}\big)\big)^2} \\ = \frac{R}{2}\sqrt{2-2\cos\big(2\pi\frac{n-1}{n}\big)} =\frac{R}{2}\sqrt{2\Big(1-\cos\big(2\pi\frac{n-1}{n}\big)\Big)}\\ =\frac{R}{2}\sqrt{4\sin\big(2\pi\frac{n-1}{n}\big)^2}=R\sin\big(\pi\frac{n-1}{n}\big)=R\sin\big(\frac{\pi}{n}\big).$$

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  • $\begingroup$ what happens if n=100000 ? The cosine is 1 and the quantity under the sqrt is negative – Georgy 3 secs ago edit $\endgroup$
    – Georgy
    Jun 17, 2015 at 7:14
  • $\begingroup$ @Georgy you are right thank you, there was a typo in my computation. I corrected it. $\endgroup$
    – Surb
    Jun 17, 2015 at 8:30

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