radius of nth circle inside a circle of given radius A circle of radius 100 cm is given. The goal is to place 7 circles of unequal radius inside the initial circle, so that each 3 of circles don't overlap each other and all of them stay inside the initial circle.
The first smaller circle is of radius 50 cm and so on.
I thought it has to do something with Ford circles, but answer r = 14.36 cm for the 7th circle is incorrect.

 A: Hint:

One way is to consider
this construction as a part of the (infinite)
Steiner’s chain of circles defined by two circles,
$\bigcirc O((0,0), R)$ and $\bigcirc o((-r,0),r)$,
with the distance between their centers $d=r$,
where $R=100,\ r=50$.
The other $6$  circles $\bigcirc O_0,\dots,\bigcirc O_5$
are sandwiched between the two,
starting with $\bigcirc O_0((r,0),r)$.
All tangent points of the kissing circles in the chain
are located on the circle (shown green in the picture) centered at $X$ with radius
$r_x=|XO_i|$, $i=0,\dots,5$,
\begin{align} 
r_x&=
\sqrt{rR-\frac{rRd^2}{(r+R)^2}}
=\frac{200}3\approx 66.7
,
\end{align}
while all the centers $O_i$, $i=0,\dots,5$,
belong to the ellipse (shown yellow in the picture)
focused at $o,\,O$.

Edit:
Using Descartes' theorem,
for the curvatures
$k_1,k_2,k_3,k_k$
of four mutually tangent circles,
\begin{align} 
k_4&=k_1+k_2+k_3\pm 2\sqrt{k_1 k_2+k_2 k_3+k_3 k_1}
\tag{1}\label{1}
.
\end{align}
In this particular arrangement of circles,
\begin{align} 
k_1&=-\tfrac1R
,\quad k_2=\tfrac2R
,\quad k_3=k_2=\tfrac2R
\end{align}
since each next circle in the chain is smaller,
the sign in \eqref{1} would be always positive
and we have
\begin{align}
k_4&=\tfrac3R
,\\
k_5&=\tfrac{6}R
,\\
k_6&=\tfrac{11}R
,\\
k_7&=\tfrac{18}R
,\\
k_8&=\tfrac{27}R
,
\end{align}
so the radii of
$\bigcirc O_i$, $i=1,\dots,5$
can be found as
\begin{align}
r_i&=\frac R{i^2+2}
.
\end{align}
For example, the seventh inner circle, the smallest one,
must have the radius
\begin{align}
r_5&=\frac R{27}
\approx 3.7037
.
\end{align}
