Let $P := (a \cos\phi, b \sin\phi)$ on an origin-centered ellipse with radii $a$ and $b$; define $c := \sqrt{a^2-b^2}$, so that the ellipse's eccentricity is $e := c/a$. The line through $P$, normal to the ellipse —that is, in direction $(b\cos\phi,a\sin\phi)$— meets the $x$-axis at $K:= (k,0)$, where $k:= c^2/a \cos\phi$. So, $K$ is the center of a circle internally tangent to the ellipse at $P$, and its radius, $r$, is given by
$$r^2 = |PK|^2 = \frac{b^2(a^2-c^2\cos^2\phi)}{a^2} = \frac{b^2(c^2-k^2)}{c^2} \tag{1}$$
so that
$$\frac{r^2}{b^2}+\frac{k^2}{c^2}=1 \tag{2}$$
This allow us to write, for some $\theta$,
$$r = b\sin\theta \qquad k = c \cos\theta \tag{3}$$
Now, suppose $\bigcirc K_0$ and $\bigcirc K_1$ are circles internally tangent to the ellipse, with respective centers and radii given by $(3)$ for $\theta = \theta_0$ and $\theta=\theta_1$. If these circles are tangent to each other (with $K_1$ "on the right" of $K_0$), then
$$\begin{align}
k_0 + r_0 &= k_1 - r_1 \\[4pt]
\to\quad -2 c \sin\frac{\theta_0 + \theta_1}{2} \sin\frac{\theta_0 - \theta_1}{2} &= -2 b \sin\frac{\theta_0 + \theta_1}{2} \cos\frac{\theta_0 - \theta_1}{2} \\[6pt]
\to\quad \tan\frac{\theta_0 - \theta_1}{2} &= \frac{b}{c} \\[6pt]
\to\quad \theta_1 &= \theta_0 - 2\arctan\frac{b}{c} \\[6pt]
&= \theta_0 - 2\arccos e \tag{4}
\end{align}$$
More generally, if circles $\bigcirc K_i$, defined by $\theta = \theta_i$ in $(3)$, form a tangent chain, then
$$\theta_i = \theta_0 - 2 i \arccos e \tag{5}$$
where index $i$ is subject to certain viability conditions (eg, $\theta_i \geq 0$) that we'll assume hold. Thus, defining $\varepsilon := 2\arccos e$, we have
$$\begin{align}
\frac{r_{i+j} + r_{i-j}}{r_i} &= \frac{b\sin(\theta_0-(i+j)\psi)+b\sin(\theta_0-(i-j)\varepsilon)}{b \sin(\theta_0-i\varepsilon)} \\[6pt]
&= 2\cos j\varepsilon = 2\cos( 2j \arccos e ) \\[4pt]
&= 2\,T_{2j}(e) \tag{6}
\end{align}$$
where $T_{2j}$ is the $2j$-th Chebyshev polynomial of the first kind. Notably, the value of $(6)$ is independent of $i$. In particular, if we take $j=3$ and both $i=4$ and $i=7$, we can write
$$\frac{r_{4-3}+r_{4+3}}{r_4} = 2\;T_{2\cdot 3}(e) =\frac{r_{7-3}+r_{7+3}}{r_7} \tag{7}$$
which gives the result. $\square$
Addendum. In this follow-up question, @g.kov asks when an ellipse allows a "perfect packing" of $n$ tangent circles along its axis. It seems reasonable to append here a justification of the condition given there.
In a perfect packing, the first and last circles in a chain are tangent to the ellipse at the endpoints of the axis, so that their radii match the ellipse's radius of curvature (namely, $b^2/a$) at those points. Thus, we have
$$r_0 = r_{n-1} = \frac{b^2}{a} \quad\to\quad \sin\theta_0 = \sin\theta_{n-1} = \frac{b}{a} \quad\to\quad \cos\theta_0 = \cos\theta_{n-1} = e \tag{8}$$
We can say that $\theta_0 = \pi - \arccos e$ and $\theta_{n-1} = \arccos e$. By $(5)$, this implies
$$\arccos e = \theta_{n-1} = \theta_0 - 2(n-1)\arccos e = (\pi - \arccos e) - 2(n-1)\arccos e \tag{9}$$
so that
$$\pi = 2n\arccos e \qquad\to\qquad \cos \frac{\pi}{2n} = e \tag{10}$$
This is equivalent to @g.kov's condition for a perfectly-packable ellipse. $\square$
Here is a solution from one of my coworkers.
Assume $AD=DE$ and $OF=KG=r$. We will prove that $HL=r$.
Define $\alpha=\angle{DAE}$. Simple angle chasing gives the other angles shown in the diagram.
$$AT=CO+OE+EK$$
$$\frac1r(OE+CO)=\frac1r(AT-EK)$$
$$\cot{(60^0-\alpha)}+\sqrt3=\cot{\left(30^0-\frac{\alpha}{2}\right)}-\cot{\left(30^0+\frac{\alpha}{2}\right)}$$
$$\frac{\cos{\alpha+\sqrt3 \sin{\alpha}}}{\sqrt3 \cos{\alpha}-\sin{\alpha}}+\sqrt3=\frac{\sin{(30^0+\alpha/2)\cos{(30^0-\alpha/2)-\cos{(30^0+\alpha/2)}\sin{(30^0-\alpha/2)}}}}{\sin{(30^0-\alpha/2)\sin{(30^0+\alpha/2)}}}$$
$$\frac{\cos{\alpha}+\sqrt3 \sin{\alpha}}{\sqrt3\cos{\alpha}-\sin{\alpha}}+\sqrt3=\frac{2\sin{\alpha}}{\cos{\alpha}-\cos{60^0}}$$
$$2\cos^2{\alpha}-\cos{\alpha}=\sqrt3 \sin{\alpha}\cos{\alpha}-\sin^2{\alpha}$$
$$\cos^2{\alpha}-\cos{\alpha}+1=\sqrt3\sin{\alpha}\cos{\alpha}$$
$$(\cos^2{\alpha}-\cos{\alpha}+1)^2=3(1-\cos^2{\alpha})\cos^2{\alpha}$$
$$(2\cos{\alpha}-1)(2\cos^3{\alpha}-1)=0$$
Obviously, $\cos{\alpha}\ne1/2$.
Let $m=2^{1/3}$.
$$\cos{\alpha}=\frac1m=\frac{1-\tan^2{(\alpha/2)}}{1+\tan^2{(\alpha/2)}}$$
$$\tan{\alpha}=\sqrt{m^2-1}$$
$$\tan{\frac{\alpha}{2}}=\sqrt{\frac{m-1}{m+1}}$$
$$\begin{align}
HL&=DE\cos{\alpha}\tan{\frac{\alpha}{2}}\\
&=(DM+ME)\cos{\alpha}\tan{\frac{\alpha}{2}}\\
&=r(\cot{\alpha}+\cot{(60^0-\alpha)})\cos\alpha\tan{\frac{\alpha}{2}}\\
&=r\left(\frac{1}{\sqrt{m^2-1}}+\frac{1+\sqrt3\sqrt{m^2-1}}{\sqrt3-\sqrt{m^2-1}}\right)\frac1m\sqrt{\frac{m-1}{m+1}}\\
&=\frac{\sqrt3mr}{(m+1)(\sqrt3-\sqrt{m^2-1})}\\
&=\frac{\sqrt3mr}{\sqrt3m+\sqrt3-(m+1)\sqrt{m^2-1}}\\
&=\frac{\sqrt3mr}{\sqrt3m+\sqrt3-\sqrt{\color{red}{(m^2-1)(m+1)^2}}}\end{align}$$
$$\color{red}{(m^2-1)(m+1)^2}=(m+2)(m^3-2)+3=0+3=3$$
$$\therefore HL=r$$
Best Answer
In rectangular coordinates $(x, y),$ the circles are $x²+y²=144$ and $(x-25)²+y²=25.$
If $y-mx-c=0$ is a common tangent, then the distance between the center of each circle and the line is the radius of the circle, so that $|c|=12√(1+m²)$ and $|25m+c|=5√(1+m²),$ so
$$5|c|=12|25m+c|.$$
There are two tangents we require, those which intersect (the radii are different) outside the segment joining the centres of the circles. By symmetry, we focus on the one with $c>0,$ so
$$5c=12|25m+c|.$$
Also, we must have that $m<0$ (the larger circle is on the left) and $-c/m>30$ (the $x$-intercept of the tangent required is on the right of the smaller circle).
There are two cases:
$5c=12(25m+c),$ so $-7c=300m,$ or
$5c=-12(25m+c),$ so $17c=-300m.$
Therefore we have that $-7c=300m,$ since it satisfies $-c/m>30.$
Therefore $c²=144(1+m²)$ and $49c²=90000m²,$ so
$90000m²/49=144+144m²,$ or $82944m²=7056,$ so
$m=-7/24,$ and $c=12.5,$ so
$$y=-7x/24+12.5,$$ intersecting the $x$-axis at $(300/7, 0).$
The acute angle of slope is $\arctan(7/24).$
The length of the tangent is given by
$$√((300/7)²-144)-√((300/7-25)²-25)=24.$$
The relevant upper arc of the large circle has central angle $π/2+\arctan(7/24),$ so the arc-length is
$$6π+12\arctan(7/24).$$
The relevant upper arc of the small circle has central angle $π/2-\arctan(7/24),$ so the arc-length is
$$5π/2-5\arctan(7/24).$$
Therefore the total length required is given by
$$2[6π+5π/2+12\arctan(7/24)-5\arctan(7/24)+24]=48+17π+14\arctan(7/24).$$
Here is a synthetic approach.
The green quadrilateral is a parallelogram and it is surmounted by a right triangle. All the action takes place in this triangle.
First, we find the length of the common tangent. The blue length in the triangle is $12-5=7$ and the green length is equal to the distance between the centres of the circles, so is $25.$ Therefore by Pythagoras the required length is given by $√(25²-7²)=24.$
To find the red arcs, we find the angle $\theta,$ whose trigonometric tangent is $24/7,$ so that $\theta=\arctan(24/7).$ But this angle corresponds to the central angle of the smaller red arc, so that the length of this arc is $5\arctan(24/7).$
Also, the central angle of the larger red arc is $π-\arctan(24/7),$ so that the length of the arc is $12π-12\arctan(24/7).$
Therefore the length of the belt required is given by
$$2[24+5\arctan(24/7)+12π-12\arctan(24/7)]=48+24π-14\arctan(24/7).$$
To see that this is the same result as above, use the identity $\arctan(y/x)+\arctan(x/y)=π/2.$