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:
With[{p = 1, e = 1/Sqrt[2], n = 31},
Animate[PolarPlot[{p/(1 - e*Cos[t]), p}, {t, -Pi, Pi},
Epilog -> {{Green,
Circle[{(p/2)*(1 - 1/(1 + e)), 0},
(p/2)*(1 + 1/(1 + e))]},
{Red,
Circle[{(e*p*Cos[th]^2)/(1 - (e*Cos[th])^2),
(e*p*Cos[th]*Sin[th])/(1 - (e*Cos[th])^2)},
p/(1 - (e*Cos[th])^2)],
Line[{{p*Cos[th]/(1 - e*Cos[th]),
p*Sin[th]/(1 - e*Cos[th])},
{-p*Cos[th]/(1 + e*Cos[th]),
-p*Sin[th]/(1 + e*Cos[th])}}]}},
Frame -> True], {th, 0, 2 Pi, 2 Pi/(n - 1)}]]
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.
Expanding a comment ...
As shown in OP's second figure: From $P$ on the hyperbola, drop a perpendicular to $M$ on the transverse axis, $Q$ be one of the points for which $\overline{MQ}$ is tangent to the circle. (We'll discuss which one of the points below.) Then $P$ and $Q$ are "corresponding points". (So, we've traded "transfer $M$ perpendicularly to the circle" in the ellipse case to "transfer $M$ tangentially to the circle" in the hyperbola case, which makes some sense in a "pole and polar" context.)
The construction can be reversed: From $Q$ on the circle, let $M$ be such that $\overline{QM}$ is tangent to the circle, then let $P$ be one of the points on the hyperbola such that $\overline{MP}$ is perpendicular to the hyperbola's transverse axis. (Again, there's ambiguity in the choice of $P$.)
Ambiguities aside, we find that every finite point $P$ on either branch of the hyperbola corresponds to some point on the unit circle except its top- and bottom-most points. The two "points at infinity" on the hyperbola correspond to those last two points on the circle.
As for those ambiguities ... This animation shows the "natural" way of resolving them. As $Q$ travels normally around the circle through Quadrants 1, 2, 3, 4, the corresponding $P$ travels along the hyperbola in Quadrants 1, 3, 2, 4; Quadrants 2 and 3 are "flipped".
This is because as $Q$ passes from Q1 to Q2 through the top-most point of the circle, $P$ passes from Q1 to Q3 "via the blue asymptote". Likewise, as $Q$ passes from Q3 to Q4, $P$ passes from Q2 to Q4 "via the red asymptote".
This quadrant-flipping notion happens to arise naturally from the equations, too. Let the hyperbola have equation
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \tag{1}$$
so that the auxiliary circle's equation is
$$x^2+y^2=a^2 \tag{2}$$
For a point $Q = (x_Q,y_Q)$ on the circle, one can show that $M = (a^2/x_Q,0)$. Of course, $P$ shares its $x$-coordinate with $M$; the $y$-coordinate, solved-for in $(1)$ has a sign ambiguity:
$$\begin{align}P &= \left(\frac{a^2}{x_Q}, \pm b \sqrt{\frac{(a^2/x_Q)^2}{a^2}-1}\right) = \left(\frac{a^2}{x_Q},\pm b\sqrt{\frac{a^2-x_Q^2}{x_Q^2}}\right)
= \left(\frac{a^2}{x_Q},\pm b\sqrt{\frac{y_Q^2}{x_Q^2}}\right) \\[4pt]
&= \left(\frac{a^2}{x_Q},\pm b\left| \frac{y_Q}{x_Q}\right|\right)\tag{3}\end{align}$$
So, we strip $y_Q/x_Q$ of its sign, only to immediately apply an ambiguous sign. That seems somewhat silly. "Quadrant-flipping" arises by letting $y_Q/x_Q$ determine its own fate, so that we have
$$P = \left(\frac{a^2}{x_Q},b\frac{y_Q}{x_Q} \right) \tag{4} $$
Thus, $P$'s $y$-coordinate is positive when $Q$'s coordinates have the same sign; that is, $P$ is in Quadrants 1 and 2 when $Q$ is in Quadrants 1 and 3; similarly, $P$ is in Quadrants 3 and 4 when $Q$ is in Quadrants 2 and 4. Again, Quadrants 2 and 3 are "flipped" for $P$ and $Q$.
Best Answer
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.