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.