I have a circle with $n$ equal tangent circles inscribed, what is the diameter of the largest circles that could be inscribed in the space left? Hello I know that the question is very long and maybe confused, I'll do my best to explain it step by step.
The problem is relevant in the cable design industry. The first step is to define the diameter of a circle and its numerosity n: from that I calculate the diameter of a supercicle that contains n tangent circles. To calculate that I cheated a little bit: I used a factor called "lay up factor" (that factor it's defined in an international ISO norm), so for instance if I have $3$ circles of diameter d the diameter s of the supercircle will be $2.13*d$. After that I have to calculate the diameter of the largest circle that can be inscribed between the supercircle and the base circles.
Maybe some images will make the problem clearer.
My final goal is to calculate the diameter of the red circles:

First step is to put n circles of diameter d tangent to each other:

I draw the supercicle (yellow) and calculate its diameter s:

It would be nice to understand how to calculate the red circles diameter for any numerosity, but I'm interested in the case of $3$ (so $3$ base circles).
 A: For $3$ mutually tangent base circles of radius $r$, their centers will form an equilateral triangle with side length $2r$.  So the distance between the center of each of the three circles and the center of this equilateral triangle is
$ d = \dfrac{2}{3} \bigg( \dfrac{\sqrt{3}}{2} (2 r) \bigg) = \dfrac{2}{\sqrt{3}} r $
The radius of the big circle is just
$ R = d + r = \bigg(1 + \dfrac{2}{\sqrt{3}} \bigg) r $
Now using Descartes' kissing circles theorem, the radius of the red circles is
$u$ and satisfies,
$ \dfrac{1}{u} = \dfrac{1}{r} + \dfrac{1}{r} - \dfrac{1}{R} + 2 \sqrt{ \dfrac{1}{r^2} - \dfrac{1}{r R} - \dfrac{1 }{rR} } $
$ \dfrac{1}{u} = \bigg(\dfrac{1}{r}\bigg) \bigg( 2 - \dfrac{\sqrt{3}}{ \sqrt{3} +2 } + 2 \sqrt{ \dfrac{2 - \sqrt{3}}{2 +  \sqrt{3} } } \bigg) $
Decimally, this evaluates to
$ \dfrac{1}{u} = \dfrac{2.07179677}{r} $
From which,
$ u = \dfrac{r}{2.07179677} = 0.482673 r $
A: Let's put the diagram on the coordinate plane, with the supercircle centered at the origin, and the $+x$-axis going through the tangent point of two blue circles and the center of one red circle. Call $S$ the radius of the supercircle, $R$ the radius of the blue circles, and $r$ the radius of the red circle.
All the blue circles have centers on a circle around the origin with radius $S-R$. So the two blue circles touching the $+x$-axis have centers at $((S-R) \cos \frac{\pi}{n}, \pm (S-R) \sin \frac{\pi}{n})$. But the distance of either center from the $x$-axis is also $R$, so if $n \geq 3$,
$$ (S-R) \sin \frac{\pi}{n} = R $$
$$ R = \frac{\sin \frac{\pi}{n}}{1+\sin \frac{\pi}{n}} S $$
$$ S = R\left(1+\csc \frac{\pi}{n}\right) $$
And we can rewrite the centers of the two blue circles touching the $+x$-axis as $(R \cot \frac{\pi}{n}, \pm R)$.
The center of the red circle on the $+x$-axis is $(S-r,0)$. The distance between the center of this red circle and the center of a tangent blue circle must be $R+r$:
$$ \sqrt{\left(S-r- R \cot \frac{\pi}{n}\right)^2 + R^2} = R+r $$
$$ \left(R + R \csc \frac{\pi}{n} - R \cot \frac{\pi}{n} - r\right)^2 + R^2 = R^2+2Rr+r^2 $$
$$ \left(R + R \tan \frac{\pi}{2n} -r\right)^2 = 2Rr + r^2 $$
$$ R^2 \left(1+ \tan \frac{\pi}{2n}\right)^2 - 2Rr \left(1+\tan\frac{\pi}{2n}\right) + r^2 = 2Rr + r^2 $$
$$ R^2 \left(1+\tan \frac{\pi}{2n}\right)^2 = 2Rr \left(2+\tan\frac{\pi}{2n}\right) $$
$$ R \left(\cos\frac{\pi}{2n} + \sin\frac{\pi}{2n}\right)^2 = 2r \left(2\cos^2 \frac{\pi}{2n} + \cos\frac{\pi}{2n} \sin\frac{\pi}{2n}\right) $$
$$ R \left(\cos^2 \frac{\pi}{2n}+2\cos\frac{\pi}{2n}\sin\frac{\pi}{2n} + \sin^2 \frac{\pi}{2n}\right) = 2r\left(1+\cos \frac{\pi}{n}+\frac{1}{2} \sin \frac{\pi}{n}\right) $$
$$ R \left(1+\sin\frac{\pi}{n}\right) = r\left(2+2\cos \frac{\pi}{n}+\sin\frac{\pi}{n}\right) $$
$$ r = \frac{1+\sin\frac{\pi}{n}}{2+2\cos\frac{\pi}{n}+\sin\frac{\pi}{n}} R $$
$$ r = \frac{\sin \frac{\pi}{n}}{2+2\cos\frac{\pi}{n}+\sin\frac{\pi}{n}} S $$
For $n=3$, we have $\frac{S}{R} = 1+\frac{2}{\sqrt{3}} \approx 2.155$, $\frac{r}{R} = 2-\sqrt{3} \approx 0.4827$, and $\frac{r}{S} = \frac{1}{1+2\sqrt{3}} \approx 0.2240$. The result for $\frac{S}{R}$ is a bit larger than your quoted $2.13$; I'm not sure why. The result for $\frac{r}{R}$ in my notation matches the result for $\frac{u}{r}$ in the notation of @Nathalieissweetandawesome's answer.
