Finding an expression for the diameter of a circle. The figure shows two circles of radius $1$ that touch at $P$. 1]1 $T$ is a common tangent line; $C_1$ is the circle that touches $C$, $D$, and $T$; $C_2$ is the circle that touches $C$, $D$, and $C_1$; $C_3$ is the circle that touches $C$, $D$, and $C_2$. This procedure can be continued indefinitely and produces an infinite sequence of circles $\{C_n\}$. Find an expression for the diameter of $C_n$.
I'll work with the radii instead of the diameters. Obviously, any diameter is twice the corresponding radius. Now, note from the figure below that the triangle $OAB$ is rectangular at $A$.



Thus,
$$
\overline{OA}^{\,2} + \overline{AB}^{\,2} = \overline{OB}^{\,2}
$$

So,
$$
(\underbrace{1}_{\overline{OA}})^2 + (\underbrace{1 - R_1}_{\overline{AB}})^2 = (\underbrace{1 + R_1}_{\overline{OB}})^2
$$
which gives you


  $$
R_1 = \frac{1}{4}
$$


Similarly, by making $B$ be the centre of the ever-smaller circles, you should be able to see that
$$
(\underbrace{1}_{\overline{OA}})^2 + (\underbrace{1 - 2R_1 - 2R_2 - \cdots - 2R_{n-1} - R_n}_{\overline{AB}})^2 = (\underbrace{1 + R_n}_{\overline{OB}})^2
\qquad(1)
$$
for $n &gt; 1$. Now define


  $$
S_{n} \equiv \sum_{k\,=\,1}^{n}R_k
\qquad (n \ge 1)
\qquad(2)
$$


so that $(1)$ can be written as
$$
1 + (1 - 2S_{n-1} - R_n)^2 = (1 + R_n)^2
\qquad(3)
$$
for $n &gt; 1$. However,
$$
(1 - 2S_{n-1} - R_n)^2 =
1 + 4S_{n-1}^2 + R_n^2 - 4S_{n-1} - 2R_n + 4S_{n-1}\,R_n
$$
and $(3)$ gives
$$
1 + 1 + 4S_{n-1}^2 + R_n^2 - 4S_{n-1} - 2R_n + 4S_{n-1}\,R_n =
1 + 2R_n + R_n^2
$$
which can be simplified to
$$
(1 - 2S_{n-1})^2 = 4(1 - S_{n-1})R_n
$$
so


  $$
R_n = \frac{1}{4}\frac{(1 - 2S_{n-1})^2}{(1 - S_{n-1})}
\qquad (n &gt; 1)
\qquad (4)
$$


This gives $R_n$ in terms of the radii of only the circles with indices smaller than $n$ so it's very amenable to a recursive calculation. For example, we already have $R_1 = 1/4$. Then $S_1 = 1/4$ and $R_2 = 1/12$. Then $S_2 = 1/3$ and $R_3 = 1/24$. Then $S_3 = 3/8$ and $R_4 = 1/40$, and so on.

Computing a few more values in succession, it's easy to discern the pattern:


  $$
S_n = \frac{n}{2(n+1)}
\qquad\mbox{and}\qquad
R_n = \frac{1}{2n(n+1)}
\qquad (n \ge 1)
$$


That these satisfy $(2)$ and $(4)$ can be easily proven by induction on $n$.
Notice, the radius say $r_1$of circle $C_1$ is given by the formula (see derivation) $$\color{red}{r_1=\frac{ab}{(\sqrt a+\sqrt b)^2}}$$ setting radii $a=b=1$ of larger identical circles, we get  $$r_1=\frac{1\cdot1}{(1+1)^2}=\frac{1}{4}$$

Now, the radius $r_2$ of circle $C_2$ externally touching the circles $C$, $D$ &amp; $C_1$ is given by the standard formula (see derivation) 
$$\color{red}{r_2=\frac{abc}{2\sqrt{abc(a+b+c)}+(ab+bc+ca)}}$$ Setting the values of $a=1, b=1, c=r_1=\frac{1}{4}$, we get radius of circle $C_2$ as follows $$r_2=\frac{1\cdot1\cdot\frac{1}{4}}{2\sqrt{1\cdot1\cdot\frac{1}{4}\left(1+1+\frac{1}{4}\right)}+1\cdot1+1\cdot \frac{1}{4}+\frac{1}{4}\cdot 1}=\frac{1}{12}$$

Similarly, 
the radius $r_3$ of circle $C_3$ externally touching the circles $C$, $D$ &amp; $C_2$ is calculated by setting the values of $a=1, b=1, c=r_2=\frac{1}{12}$ in the above standard formula as follows $$r_3=\frac{1\cdot1\cdot\frac{1}{12}}{2\sqrt{1\cdot1\cdot\frac{1}{12}\left(1+1+\frac{1}{12}\right)}+1\cdot1+1\cdot \frac{1}{12}+\frac{1}{12}\cdot 1}=\frac{1}{24}$$ Thus, continuing the same procedure, we can calculate radius of any circle $C_n$ using above standard formula.