What is the least number of circles with radius $r$ is required to cover a circumference of a circle with radius $R>r$?

Question

I was playing around with circles and tried this: Given a circle with radius $R$ and $r<R$ What is the least number of circles with radius $r$ is required to cover the circumference of the circle with radius $R$?

I think the best possible way to do this is when the centres of the circles with radius $r$ are on the circumference of a circle with radius $R$ and are externally tangent to each other (except the one circle which overlaps) at least when $\frac{R}{r}$ is big enough (for example $R=r+ \varepsilon $ ) but how to prove that?

(The centres of the smaller circles doesn't have to be on the circumference of a circle with radius $R$ The goal is to find the least number of circles with radius $r$ is required to cover a circumference of a circle with radius $R$.

Given the above construction (putting the centre of the first small circle on the the circumference of the circle with radius $R$ and making the $n+1$th circle externally tangent to the $n$th circle with the centre on the circumference of the big circle) we get:

When $R=5, r=1$ we need $16 $ circles

When $R=6, r=1$ we need $19 $ circles

When $R=2\pi, r=1$ we need $20 $ circles

When $R=7, r=1$ we need $22 $ circles

all of the examples that I tried have overlapping circles, My initial solution is to find the some regular polygon that
the big circle is internally tangent to it but I this is very hard and it will only work when the small circles don't overlap which it seems not to happen very often.

My second question is: what is ration $\frac{R}{r}$ have to be so the small circles don't overlap ?

Answer 1

The angle subtended by the small circle of radius $r$ at the center of the big circle of radius $R$ is

$ \theta = 2 \sin^{-1} \left(\dfrac{r}{R}\right) $

Since the complete angle is $2 \pi$, then the number of small circles necessary to cover the big circle is

$ X = \dfrac{ \pi }{ \sin^{-1} \left( \dfrac{r}{R} \right) } $

If $X$ is an integer, then we're done. Otherwise the minimum number to cover the big circle will be

$ X' = \lfloor X \rfloor + 1 $

To demonstrate this formula, I've take $R = 10$, and set the number of small circles to $5$. From this I calculated $r$

From this, the angle subtended by each small circle is $\theta = \dfrac{2 \pi}{5} $

Therefore, according to the above formulas:

$ r = R \sin \left( \dfrac{ \theta }{ 2 } \right) = 10 \sin \left( \dfrac{ \pi }{5} \right) $

When I plotted this, I got this

Obviously, the small green circles do not cover the big circle completely.

As @Henry noted below the in the comments, one way to rectify this is by choosing to keep the radius of the small circles as is, but move the centers to lie on a circle of radius $R' = R \cos\left( \dfrac{\pi}{n} \right) $.

This is shown below.

Another way to handle the gaps is by modifying the original formulas as follows:

Firstly,

$ \theta = 2 \tan^{-1} \left( \dfrac{r } {R } \right) $

Secondly, the radius on which the centers of the small circles is larger than $R$. It is easy to see that it is given by

$ R' = R \sec \left( \dfrac{\theta}{2} \right) = \sqrt{R^2 + r^2} $

This modification is shown in the figure below, where the solid black circle is our big circle, and the dashed circle is the circle on which the centers of the small circles (each of radius $r$) lie. Note that $r$ in this case is different than the $r$ of the initial trial. Now the small circles cover all of the big circle which of radius $10$.

However, if you want the centers of the small circles to be on the big circle, then some overlapping will occur between neighboring circles.

It is easy, in this case, to calculate the radius of each small circle to be

$ r = R \left( 2 \sin \left( \dfrac{\theta}{4} \right) \right) $

where

$ \theta = \dfrac{2 \pi}{n} $

where $n$ is the number of circles required.

Here's the resulting figure with $n = 5$

And for $n = 16 $

Answer 2

This is for when centres lie on the circumference, and I can think of the following idea. Let's take the circumference as $2πR$. Now $l=R\theta$. We need to establish a relation between $\theta$ and $r$.

$\angle DAE=\theta$. By properties of circles $\angle EBD=180°-\frac{\theta}{2}$

Now we use law of cosines, $$ED^2=2R^2-2R^2\cos\theta=2r^2-2r^2\cos(180°-\frac{\theta}{2})$$ $$\implies \frac{R^2}{r^2}=\frac{1+\cos\frac{\theta}{2}}{1-\cos\theta}$$ Taking $\cos \frac{\theta}{2}=t$, $$R^2(1-(2t^2-1))=r^2(1+t)$$ $$\implies 2R^2(1-t)(1+t)=r^2(1+t)$$ But $t≠-1$ so $$t=1-\frac{r^2}{2R^2}\\\implies \theta =2\arccos\bigg(1-\frac{r^2}{2R^2}\bigg)$$

Hence $$l=2R\arccos\bigg(1-\frac{r^2}{2R^2}\bigg)$$ Number of circles required $=\lfloor \frac{2πR}{l}\rfloor +1$(overlapping) or $\frac{2πR}{l}$(non overlapping), depending upon the value of $l$

This matches the results you're getting.

For them not to overlap, $$\arccos\bigg(1-\frac{r^2}{2R^2}\bigg)=kπ,\frac{1}{k}\in \mathbb{N}\\ \implies 1-\cos(kπ)=\bigg(\frac{r}{\sqrt 2R}\bigg)^2\\\implies \frac{r}{R}=\sqrt{2}\sqrt{1-\cos kπ}=2|\sin\frac{kπ}{2}|$$ For example, say $k=\frac{1}{3}$, then $r=R$, and number of circles $=3$

Answer 3

4th edition

OP asks for the number of unit circles required to cover the circumference of a given circle of radius $R>1$. The following is a kind of converse approach: “Given a number of circles of radius $r=1$ covering the circumference of a circle of radius $R>1$, to determine the placement of the center of the unit circle.”

I. Let a circle with radius $OD=R$ be inscribed in a regular $n$-gon with side $AB$. Let $DF=OD$ meet $OB$ extended at $F$. Let $DE$ join adjacent tangency points $D$, $E$ and intersect $OF$ at $G$. $\angle DOB=\frac{\alpha}{2}$, where $\alpha$ is the central angle of a regular $n$-gon.

The circle with radius $FD$ is not a unit circle, since $FD=OD=R$ and $r=1<R$. But as point $F$ moves up toward $B$, $FD<R$, and when $F$ coincides with $B$, then with $BD=1$, $$OD=R=\tan^{-1}\left(\frac{\alpha}{2}\right)$$And when $F$ coincides with $G$, then with $GD=1$, $$OD=R=\sin^{-1}\left(\frac{\alpha}{2}\right)$$

Since $GD$ is radius of the smallest circle through tangent points $D$, $E$, then the circumference of a circle of radius $R$ inscribed in a regular $n$-gon can be covered by $n$ unit circles centered on a segment of the $n$-gon's extended spokes, such as $GF$, and$$1<R\le\sin^{-1}\left(\frac{\alpha}{2}\right)$$For example, if $n=6$, then$$1<R\le\sin^{-1}30^o=2$$that is, the circumference of a circle of radius$$1<R\le2$$can be covered by six unit circles. For integer $R=2$, $G$ is the center of a unit circle. For $1<R<2$ the center lies inside the hexagon on $BG$ or outside on $BF$.

The location on $GF$ of the center of the unit circle for integer values of $R$ depends on $n$. For $n=7, 8, 9$, $R=2$ for unit circles centered increasingly far out on $BF$. For $n=10, 11, 12$, again $R=3$ for circles centered increasingly far out on segment $BF$. And the same for $n=13, 14, 15$, with $R=4$, and for $n=16, 17, 18$, with $R=5$: the $n$ unit circles have their centers beyond the vertices of the circumscribed regular $n$-gon.

But as in OP’s examples, for $n=19, 22$, and again for $n=44$, $R=6, 7, 14$, respectively, for unit circles centered inside the $n$-gon, on $BG$.

[Addendum: Centers of unit circles for integer $R$ typically lie on $BF$, i.e. outside the $n$-gon, but they lie inside, on $BG$, when integer $R$ is a multiple of $7$ and $n=66, 88$, and on $BF$ but very close to $B$ for $n=110, 132, 154, 176,...$ and apparently whenever $n$ is a multiple of $22$. This brings to mind the well-known approximation$$\pi\approx\frac{22}{7}$$and merits further study.]

II. If $n$ smaller circles have their centers on the circumference of the larger circle and are externally tangent, then they cannot cover the circumference of the larger circle; points of tangency lie within the larger circle and arcs of the larger circumference between the tangent circles remain uncovered, as seen in the figure below.

III. For the $n$ circles not to overlap, it's necessary that$$\frac{R}{r}=\tan^{-1}\left(\frac{\alpha}{2}\right)$$with the unit circles centered not on the larger circumference but on the vertices of the regular $n$-gon circumscribing the larger circle. I have not yet found a unit circle centered on $B$, and hence tangent to its neighbors, for integer $R$, but of course a unit circle will be centered on $B$ for some non-integer $R$.

Answer 4

Since it's fine if the small circles overlap, it's enough to find the largest angle $\theta$ of the large circle that a small circle can cover (the diagram here might help). The number of required circles is then given by $\lceil\frac{2\pi}{\theta}\rceil$. I'll do this by looking at the coordinates.

WLOG let the large circle be given by $x^2+y^2=R^2$, and a small circle by $(x-h)^2+y^2=R^2$, where $R-r<h<R+r$ (so that the small circle does cover some part of the large circle). If you solve for where they intersect, the intersection points are given by an $x$-value of $\frac{R^2-r^2+h^2}{2h}$. This is minimized with $h=\sqrt{R^2-r^2}$, and then $x=\sqrt{R^2-r^2}$ as well. The angle is then given by $2\cos^{-1}\left(\frac{\sqrt{R^2-r^2}}{R}\right)=2\sin^{-1}\left(\frac{r}{R}\right)$. So the best for the number of circles is then given by $$\left\lceil\frac{\pi}{\sin^{-1}(r/R)}\right\rceil$$ as the other answers indicated.