Let $AE$ and $BF$ be perpendicular to $AB$, with $E$ and $F$ lying on $CD$.
Let $N$ be the reflection of $P$, across the midpoint of $EF$. It follows that $CN=CE+EN=EP+PF=NF+FD=ND$, so $N$ is the midpoint of $CD$.
It follows that $ABFE$ is a rectangle, hence it is similar to $BAEF$.
Applying power of point $M$ to both circles, it can be seen that $MA=MB=MP=\frac{1}{2}AB$.
The reflection bringing $ABFE$ to $BAEF$ maps $M$ to itself and $P$ to $N$.
Hence, $MN=MP=\frac{1}{2}AB$.
We use the point $H$ and the circle that passes through all the contact points of neighboring circles of the chain, which was mentioned by the OP in his post. By the way, $H$ is located on the segment $LB$ a distance of $\frac{a}{6}$ away from the point $L$.
Start the construction by drawing the aforementioned circle, which is marked by the two end-points $C$ and $D$ and has its center located at $H$. Furthermore, its radius is equal to $HM_{01}$. Remember that the point $M_{01}$ is already available at this moment. This circle cuts the circle $O_1$ at $M_{12}$. Draw and extend the line $O_1 M_{12}$. We know that the center of the sought circle of the chain lies on this line.
Now, draw the two lines $M_{12}H$ and $O_1L$. The line $O_1L$ passes through the contact point $Q_1$ of the blue and yellow circles. Then, draw a line perpendicular to the $O_1L$ at $Q_1$ to intersect $M_{12}H$ at $N$. Construct an auxiliary circle with radius $Q_1N$ and having its center at $N$. This circle meets the blue circle at $Q_2$. The line $Q_2N$ is the common tangent of the blue circle and the sought member of the circle chain. Furthermore, $Q_2$ is the contact point of these two circles. Therefore, the center of the sought circle lies on the extended portion of the line $Q_2L$.
Now, we have two lines harboring the center of the sought circle, i.e. $O_1 M_{12}$ and $Q_2L$. Therefore, the point $O_2$, where these two lines meet, is the center of the next member of the circle chain. To complete the construction, draw the circle with radius $O_2Q_2$ or $O_2M_{12}$ taking $O_2$ as its center.
$\underline{\mathrm{Added\space at\space OP’s\space Request\space …}}$
A geometric configuration, in which three circles (e.g. green, yellow and blue circles) touch each other externally, has a unique point (in our case $N$), where the three common tangents coincide. Therefore, we can draw any two of the three common tangents to obtain this point. In other words, the point of intersection of any two common tangents (e.g. $NM_{12}$ and $NQ_1$) gives us this point. That is how we obtained the point $N$ in the first place. Using $Euclid\space Theorem\space 59$, we can show that the three distance from this point to each contact point of a pair of circles are equal. That is why we constructed an auxiliary circle with radius $Q_1N$ (or $NM_{12}$) and having its center at $N$. This circle cuts the blue circle at $Q_2$ giving us the third common tangent $NQ_2$. Now, we know that there exist a unique circle, which touches the yellow circle at $M_{12}$ and the blue circle at $Q_2$ externally. The center of this circle lies at the point of intersection of the two lines $O_1 M_{12}$ and $Q_2L$. The radius of this green circle is chosen as $O_2 Q_2$ to let its circumference pass through the point $Q_2$.
However, the story does not end here, because we have not yet explicitly stated that the green circle we obtained touches the red circle internally. This can be proven using trigonometry. But first, we would like to put forward the following argument. If the green circle cuts or fail to touch the red circle, then, this problem has no solution, because you cannot find another circle that touches the red circle internally and the blue circles externally while touching the yellow circle at $M_{12}$ externally.
If you want us to post the proof, please let us know
Best Answer
Because of triangle similarity we have $${CO'\over CO} ={R\over r}\implies \boxed{CO = OO'{r\over R-r}}$$
Let $C_1$ be a center oh homothety which takes $c_2$ to $c_1$ and let $C_2$ be a center oh homothety which takes $c_2$ to $c_3$. All you need to prove is $C_1=C_2$ using boxed formula.