Numbers of circles around a circle

Imagine there are $n \geq 3$ circles surrounding your unit circle. Then the situation would look like:

Whence $\cos(x)=\frac{r}{r+1}$. The angle $x$ is half of the interior angle of the corresponding regular polygon, so $x=\frac{n-2}{2n} \cdot 180^\circ$. You can then solve for $r=\dfrac{\cos(x)}{1-\cos(x)}$.

For example

• when $n=4$ we have $r=\dfrac{\cos(45^\circ)}{1-\cos(45^\circ)}=1+\sqrt{2}=2.41421\ldots$.
• when $n=6$ we have $r=\dfrac{\cos(60^\circ)}{1-\cos(60^\circ)}=1$.
• when $n=8$ we have $r=\dfrac{\cos(67.5^\circ)}{1-\cos(67.5^\circ)}=0.619914\ldots$.

Let's recollect a very simple fact about the situation when having n identical circles, $n\geq3$ perfectly surrounding the unit circle. If we connect all intersection points of these circles with the unit circle we get a regular polygon inscribed to the unit circle. The side length formula of the regular polygon inscribed to a circle is $l=2r\sin(\frac{\pi}{n})$, $n$ - the number of the sides of the polygon and is equal to the number of the surrounding circles. In our case $l=2\sin(\frac{\pi}{n})$. Now imagine that in all the picture you have there remains only the unit circle and two of the surrounding circles that are against each other. Connect the three centers of these circles and also consider the side of the polygon inside to the unit circle and get similar triangles that by Thales theorem yield:

$$ \frac{l}{2r}=\frac{1}{1+r};r=\frac{2\sin(\frac{\pi}{n})}{2-2\sin(\frac{\pi}{n})}=\frac{\sin(\frac{\pi}{n})}{1-\sin(\frac{\pi}{n})}.$$

The proof is complete.