# $N$ circles on a circle

Perhaps a rather elementary question, but I simply couldn't figure out the calculations on this one. Say one takes a circle centered at the origin with radius $R$. He or she then proceeds to place $N$ circles with radius $r$ ($R > r$) on the larger circles circumference equidistantly, so every $2 \pi / N$ in the angular sense. What is then the relationship between $R$ and $r$ such that all neighboring circles exactly touch?

I've been trying to write down some equations with arc lengths and such for $N = 4$, but I can't seem to get anything sensible out of it.

• When you say "place N circles on the larger circle", do you mean that the smaller circles' centers are on the larger circle's circumference? – weux082690 Jul 2 '15 at 15:42
• oh, that's a good question. I assumed that we were placing circles around the inner circle, not on the inner circles boundary. – davidlowryduda Jul 2 '15 at 15:45
• Yeah, that is a very good point. I apologise for that, sometimes it's hard when you have a very clear picture in your head. What I should have said is indeed placing the smaller circle centers on the larger circles circumference. – user3183724 Jul 2 '15 at 15:49

If two circles are tangent (just touch) then the distance between their centers is the sum of their radii, in your case $2r$. Since these centers are also on the circumference of the larger circle, they are distance $R$ from the center of the large circle. This makes a triangle between the three centers, in which two sides are known ($R$ and $R$ from the center of the large circle the the centers of the smaller circles) and the angle between them ($2 \pi/N$ being the angular distance between the smaller circles) and so using the law of cosines one can solve for the third side (which will be $2r$, twice the smaller circle's radius).

Doing the math, I get that $r/R$ is $\sqrt{(1- \cos \alpha)/2}$ where $\alpha$ is $2 \pi/N$.

This is an answer to the similar question: what if the outer circles are placed on the boundary of the inner circle?

I still find it easier to think about the inverse problem. Keeping up with the idea as before (in my earlier answer), the four centers of the outer circles will form a square. The inner circle is the unique circle going through the centers of these four outer circles. A quick drawing reveals that the square formed by the four outer circles' centers is inscribed within the inner circle.

Therefore the diagonal of the square is a diameter of the inner circle. The diagonal of the square has length $2r\sqrt 2$, and so the diameter of the inner circle is $2r \sqrt 2$. Thus the radius $R$ of the inner circle is $r \sqrt 2$, and we see the relationship $\frac{R}{r} = \sqrt 2$.

• Yes, this makes perfect sense. Thanks a lot. – user3183724 Jul 2 '15 at 16:25
• Actually, let me change that. The way you have done it now, is this not only valid for the case of $N = 4$? – user3183724 Jul 2 '15 at 16:26
• It very easily generalizes. The generalization leads you to weux's answer. – davidlowryduda Jul 2 '15 at 16:56
• @mixedmath I think what you have suggested is only a special case when N = 4h; for some positive integral values of h. First of all, it needs 4 circles symmetrically situated on both sides of the axes. For any adjacent two, one can stuck k circles (if possible) in between making a total of 4(1 + k) = N. – Mick Jul 3 '15 at 5:17
• I think you are overly focused on diameters. The natural generalization is to use interior angles of regular n-gons. You know the angles, you know the side-length of the n-gon, so you know all the geometry. – davidlowryduda Jul 3 '15 at 5:19

I find thinking about the inverse problem to be more intuitive. Let's start with $4$ circles of radius $r$ whose centers are arranged in a square and which are placed so that each circle touches its two neighbors.

From this perspective, it's quite clear that the sidelength of the square is $2r$. If we were to place a circle in the center of the four circles, it would also be in the center of the square. It's easy to see that the center of the square is $r\sqrt 2$ away from the centers of the four circles with a drawn picture.

So the inner circle must have radius $R = r\sqrt 2 - r = r(\sqrt 2 - 1)$.

In other words, we must have that $\frac{R}{r} = \sqrt{2} - 1$.

• I'm so very sorry for being too vague, but this is sadly not the question I meant to ask. – user3183724 Jul 2 '15 at 15:50