To show the center of homothety of the biggest and smallest circle lies in the common tangent over T $c_1$ centered at $A$ passing through $B$.
$BB′$ is a diameter of $c_1$.
$T$ a random point in segment $BB′$.
$c_2$ centered at $B′$ passing through $T$.
$c_3$ centered at $B$ passing through $T$.
$c_4$ tangent externally to $c_2$ and $c_3$ and internally tangent to $c_1$
$F$ is center of $c_4$ and $H,I$ are tangency points.

It is clear to me that $Z = HI \cap AF$ is the second homothety center of $c_1$ and $c_4$ and I would like to prove that it also lies in that line perpendicular to $AB$ through $T$.
important related result that you probably should know: Show these three circles share their external common tangent lines
This seems to be a general result about soddy circles
 A: *

*Let common tangent at $T$ meet $AF$ at $Y$ and let perpendicular to $AB$ through $F$ meet $AB$ at $L$.  Then we calculate $y=LT$ by Pythagoras theorem:
$$ B'F^2-B'L^2 = LF^2 =BF^2-BL^2$$
so $$ (b+c)^2-(b-y)^2 = (2a+b+c)^2-(2a+b+y)^2$$
and so we get $$y= {ac\over a+b}$$ so $${AY\over FY}  = {AT\over LT} = {a\over y} = {a+b\over c}$$


*On the other hand let $X$ be in $HI\cap AF$.

Homothety $H_1$ at $H$ and coefficient ${b\over c}$ takes $F$ to $B'$ and homothety $H_2$ at $G$ and coefficient ${a+b\over b}$ takes $B'$ to $A$, so composition $H_2\circ H_1$ takes $F$ to $A$ and has center at $FA\cap GH =X$. This composition has coefficent $${a+b\over b}\cdot {b\over c} = {a+b\over c}$$ so $X$ divides $AF$ in the same ratio as $Y$ and thus $X=Y$ and we are done.
A: The argument in Aqua's answer can be shortened as follows.  We use the same point names, but here $a,b,c$ are the radii of the circles centered at $A,B',F$ respectively (this changes the meaning of $a$).  Let $LT:TA$ be $x$.
As described in Yiu's Triangle Geometry, pg 2, the internal homothetic center $X$ (aka internal center of similitude) of two circles $O(R),I(r)$ divides the segment $OI$ in the ratio $R:r$.  Thus the internal homothetic point of $F(c),A(a)$ divides $FA$ in the ratio $c:a$.
Using Pythagoras' Theorem as in Aqua's answer we get
$$
(b+c)^2-(b-x(a-b))^2=(2a=b+c)^2-(2a-b+x(a-b))^2
 $$
Solving for $x$ (using an online solver if we are lazy) we get $x=\dfrac{c}{a}$.  Thus
$$
FY:YA  = LT:TA = x = c:a,
$$
so $Y$ is the internal homothetic center of $c_1,c_4$.
