N Circle touching x-axis and fixed point If a circle $C_0$, with radius 1 unit touches both the axes and as well as line ($ L_1
$) through P(0,4), $L_1$
cut the x-axis at $(x_1 ,0)$ . Again a circle $C_1$ is drawn touching x-axis, line $L_1$
 and another line $L_2$
through point P. $L_2$
 intersects x-axis at $(  x_2,0)$ and this process is repeated n times. Then the value of : $$\lim_{n\to \infty }\frac{x_n}{2^n} $$ is
I am only able to find $C_1$ which is $(x-1)^2+(y-1)^2=1$ and $L_1$ which is 3x+4y=12 but not able to proceed forward
 A: Let points $C_n=(c_n,1)$ be the centers of the circles and $X_{n}=(x_n,0)$ the intersections of tangent lines with $x$-axis. As $C_{n+1}X_nC_n$ is a right triangle (because $C_{n+1}X_n$ and $X_nC_n$ are bisectors of adjacent angles), if we set $\delta_n=c_{n+1}-x_n$ then we have (by the geometric mean theorem)
$x_n-c_n=1/\delta_n$ and 
$$
\tag{1}
c_{n+1}-c_n=1/\delta_n+\delta_n.
$$
On the other hand, if $T_n$ is the tangency point in figure below, we have 
$PT_n=PX_n-\delta_n$ and (by the Pythagorean theorem) $PT_n^2=PC_n^2-1$, that is:
$$
\Big(\sqrt{(c_{n+1}-\delta_n)^2+16}-\delta_n\Big)^2=c_{n+1}^2+9-1,
$$
which gives:
$$
\tag{2}
c_{n+1}={2\over\delta_n}-\delta_n.
$$
Plugging this into $(1)$ gives then
$$
c_{n}={1\over\delta_n}-2\delta_n
\quad\text{and}\quad
{1\over\delta_{n+1}}-2\delta_{n+1}={2\over\delta_n}-\delta_n,
$$
that is:
$$
\delta_{n+1}={1\over2}\delta_n.
$$
As we can easily compute $\delta_1={1\over2}$ it follows that
$$
\delta_n={1\over2^n}
\quad\text{and}\quad
x_n={2\over\delta_n}-2\delta_n=2^{n+1}-{1\over2^{n-1}},
$$
whence $\displaystyle\lim_{n\to\infty}{x_n\over 2^n}=2$.

A: 
This is a known configuration 
when $\triangle POX_n$ with the height $h=|OP|$ and inradius $r$
is split into $n$
triangles $\triangle POX_1,\triangle PX_1X_2,\dots\triangle PX_{n-1}X_n$,
such that all their incircles have the same radius $r_n$.
The radius $r$ of the inscribed circle of $\triangle POX_n$
is found as
\begin{align} 
r&=\frac h2\,\left(1- \left(1-\frac{2r_n}h \right)^n \right)
\tag{1}\label{1}
.
\end{align}
For $h=4,r_n=1$ we have 
\begin{align} 
r&=2-2^{1-n}
\tag{2}\label{2}
.
\end{align}
\begin{align}
2r&=|OP|+|OX_n|-|PX_n|
=h+x_n-\sqrt{h^2+x_n^2}
,\\ 
x_n&=
\frac{2r(h-r)}{h-2r}
=2^{n+1}-2^{1-n}
,\\
\lim_{n\to \infty }\frac{x_n}{2^n}
&=\frac{2^{n+1}-2^{1-n}}{2^n}
=2
.
\end{align}
