How to find the perimeter of a triangle which connects three circles? The figure from below show three circles, T and P are tangential points and A, B and C are the center of each circle. How can I find the perimeter of the triangle ABC if the distance between AC is 7 centimeters?.

Edit:
Can it be proved that the line joining the points T, B, and A is a straight line? Because the smaller circle with center at B is not in scale.
 A: Let's draw the tangent lines 
$\ell_p$ and $\ell_t$
through the points $P,T$, $\ell_p \cap \ell_t=Q$. 

Note that $\ell_p$ is tangent to both circles $B$ and $C$,
similarly, $\ell_t$ is tangent to both circles $B$ and $A$,
and $BQ$ is a bisector of $\angle PQT$.
Hence 
\begin{align}
BT&\subset AT,\quad |BT|=|BP|=r\in(0,\tfrac12R]
,\\ 
|AB|&=R-r,
\end{align}
and a variable part of the perimeter of $\triangle ABC$,
$|AB|+|BP|=R$.
Edit
The locus of all centers of the circles $B$,
that touch one of the other two circles ($A,B$)
internally, and the other one externally,
is an ellipse $\mathcal{E}$.
The smallest (degenerate) such circle with $r=0$
is centered at $H_1,H_2$,
the biggest one with $r=\tfrac12R$ is 
centered at points $G_1,G_2$, in which case 
the lines $\ell_1\parallel\ell_2$ and the point $Q$
is at the infinity.
Edit

A: 
Notice that $$\bar{AB} = \bar{AT}-\bar{BT}$$,
$$\bar{BC}=\bar{BP}+\bar{PC}$$
So the perimeter is $$P=\bar{AC}+\bar{AB}+\bar{BC}$$
Replacing,$$P=\bar{AC}+\bar{AT}-\bar{BT}+\bar{BP}+\bar{PC}$$.
Also notice that $\bar{BT}=\bar{BP}$ (radii of the same circle), $\bar{AT}=\bar{AC}=\bar{PC}$, so
$$P=3\bar{AC}$$
$$P=21cm$$
