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. $\square$
As in Blue's first comment, the limaçon of a circle with $b=0$ is indeed the circle itself. (That the pedal curve of a circle is the circle itself can be seen by using the basic foot-of-the-perpendicular equations with no hassle, too.)
The parabola's auxiliary circle, as in the second comment, can be visualised as an auxiliary circle itself; as the eccentricity of an ellipse graphed with a focus close to the origin grows closer to one, and one end shoots off into infinity, the part of the auxiliary circle touching the visible end becomes more vertical, until when $e=1$, it becomes the tangent at the vertex. This line then flips(as $e$ grows past 1) to become the auxiliary circle for the hyperbola.
Thus, the auxiliary circle can be understood as the pedal of the conic from its focus.
Best Answer
I present here a rather elaborate coordinate geometry proof. (I'd sure like to see a simpler way of going about this!)
Take the ellipse with eccentricity $\varepsilon\in(0,1)$ and semilatus rectum $p$ to have the polar equation
$$r=\frac{p}{1-\varepsilon\cos\,\theta}$$
such that one of the ellipse's foci is at the origin, and the circle whose diameter is the latus rectum has the equation $r=p$. From this configuration, we find that the associate circle corresponding to the focus at the origin has its center at $\left(\dfrac{p}{2}\left(1-\dfrac1{1+\varepsilon}\right),0\right)$ and a radius of $\dfrac{p}{2}\left(1+\dfrac1{1+\varepsilon}\right)$.
Let a focal chord corresponding to the focus at the origin be at an angle $\varphi$ from the horizontal axis. We find that the line $\theta=\varphi$ intersects the ellipse at the angle values $\theta=\varphi$ and $\theta=\varphi+\pi$, corresponding to the points
$$\left(\frac{\pm p\cos\,\varphi}{1\mp\varepsilon\cos\varphi},\frac{\pm p\sin\,\varphi}{1\mp\varepsilon\cos\,\varphi}\right)$$
(I thank Blue for noting this simplification; the previous version of this answer took a more circuitous route.)
The circle whose diameter is the segment joining these two points has its center at $\left(\dfrac{\varepsilon p\,\cos^2\varphi}{1-(\varepsilon\cos\,\varphi)^2},\dfrac{\varepsilon p\cos\,\varphi\sin\,\varphi}{1-(\varepsilon\cos\,\varphi)^2}\right)$ and a radius of $\dfrac{p}{1-(\varepsilon\cos\,\varphi)^2}$. To verify that this circle is tangent to the associate circle, we find the radical line of these two circles (which coincides with the common tangent line of two tangent circles); we find the equation of the radical line to be
$$p\left(\frac{2\varepsilon}{\sec^2\varphi-\varepsilon^2}+\frac1{1+\varepsilon}-1\right)x+\frac{2 \varepsilon p \tan\,\varphi}{\sec^2\varphi-\varepsilon^2}y+\frac{\varepsilon p^2 (\varepsilon+\sec^2\varphi)}{(1+\varepsilon)(\sec^2\varphi-\varepsilon^2)}=0$$
The point of tangency is then found to be
$$\left(\frac{p(\tan^2\varphi-\varepsilon-1)}{(\varepsilon+1)^2+\tan^2\varphi},-\frac{p(\varepsilon+2)\tan\,\varphi}{(\varepsilon+1)^2+\tan^2\varphi}\right)$$
I'll leave the verification that the line perpendicular to the radical line at the point of tangency passes through the center of the associate circle to you.
Here is a Mathematica demonstration:
Here's a parabolic version:
Some tweaking in the code given above is needed to handle the hyperbolic case
e > 1; I'll leave this as an exercise to the reader.