Circles tangent to a parabola For the past two weeks I was struggling with solving the following problem.
Description of variables:
$(x_n,y_n)$ - center point of the circle $C_n$
$r_n$ - radius of the circle $C_n$
Given the curve $k:y=x^2,x\geqslant0$, and knowing that for every $n\in\mathbb{N} \wedge  n>0$ $C_n$ is tangent to $k$, the $OX$ axis and externally tangent to $C_{n-1}$, find the explicit formula for $x_n$, $y_n$ and $r_n$ given the initial condition $C_0: (x-x_0)^2+(y-y_0)^2=r_0^2$, where $C_0$ is tangent to $OX$ and $k$.
It is obvious that $r_n=y_n$, but I did not manage to derive a sensible system of equations.
I said "sensible", because this is the only one I could think of:
$\begin{cases} \frac{x_T-x_n}{y_n-y_T}=2x_T \\ y_T=x_T^2\\\sqrt{(x_T-x_n)^2+(y_T-y_n)^2}=y_n \\ \sqrt{(x_n-x_{n-1})^2+(y_n-y_{n-1})^2}=y_n+y_{n-1} \end{cases}$
Is there no more convenient way to approach this?
Thank you for any help in advance. 
 A: This is not intended to be an answer as such but more of a suggestion as to how one might be able to proceed further. 
It seems that the equation satisfied by the coordinates of the centre $(x,y)$ of any circle which touches the parabola and the $x$ axis is itself very complicated: $$x^2=\left(\frac{8y-1+\sqrt{16y+1}}{8}\right)\left(\frac{3+\sqrt{16y+1}}{4}\right)^2$$
Despite its appearance this is a fairly innocent-looking curve passing as you would expect between the parabola $y=x^2$ and the $x$ axis.
This was obtained by just considering the requirement that each circle must touch the parabola and the $x$ axis. Therefore somehow this would need to be coupled with the requirement that the circles touch each other in series, which can be written as $$x_{n+1}=x_n+2\sqrt{y_ny_{n+1}}$$
But I don't know how to link these up. Any ideas?
A: Standard  parametric equn of parabola $ y = x^2 / (4 f) $ is
( in this case we can take $ f= \frac14 $ )
$$ x = 2 f t ; \, y = f t^2; \, t = \tan \phi $$
Radius of contacting circle $ = y  \sec\phi   = f t^2 \sqrt{1+t^2} $
Distance to right of parabola point x-coordinate is $ y\,  t = f t^3  $
So the center of circle on x-axis $ ( 2 f t + f t^3, 0) $
Rest you can proceed.
EDIT1:  
Contacting circles as shown. 
$r$ is radius of bigger contact circle given independently as in sketch.
$$ x = 2 ft + r \sin \phi = 2 f t  + \frac {r  t}{ \sqrt{1+t^2}}$$
$$ y = ft^2 - \frac{r}{ \sqrt{1+t^2} }   $$
So you can solve for  $ t =  \tan \phi $ when $ y=r $ from the resulting sixth degree transcendental equation :
$$  \frac{f\; t^2}{r}= 1 -\frac{1}{ \sqrt{1+t^2} } $$

