Quick recap, since there are several circles of Apollonius:
Given a triangle with fixed base and the other two sides in known ratio, the circle of Apollonius (of the first type) gives the possible positions of the third point.
Now constructing that circle from the two base points $A$ and $B$ and another point $D$ that divides $AB$ in the appropriate ratio is a question considered here, which can be achieved in various different ways. $D$ is clearly on the Apollonian circle, so finding one more point and knowing that the center of the circle $O$ is on $AB$ is sufficient to construct it.
The construction I gave is this:
Construct the circle $\color{#0B2}{\Gamma}$ centred on $D$ through the nearer of $A$ and $B$, which I will assume is point $B$. Also construct a line perpendicular to $AB$ at $B$. Then form the tangent to circle $\color{#0B2}{\Gamma}$ from $A$. This intersects the perpendicular at $E$, which is on the Apollonius circle because $\triangle AEB$ is similar to the right triangle $\triangle ADT$ (with $T$ the tangent point of $A$ on $\color{#0B2}{\Gamma}$), so $|AE|:|EB| = |AD|:|DT| = |AD|:|BD|$ as required.
Now the extra claim I make - visible in the diagram - is that $AE$ is a tangent to the Apollonius circle. I initially thought I had proved this via the power of point $A$ to the circle with a bit of Pythagoras, but I don't really think this works. However I do think that $AE$ is actually tangent and certainly this reliably works in interactive geometry tools.
So my question is: can you prove (or disprove) that $AE$ is tangent to the circle of Apollonius shown?
From $|AE|:|EB| = |AD|:|BD|$, $DE$ is angle bisector of $\angle AEB$.
Then $$\angle BEO = \angle DEO - \angle DEB=\angle ODE - \angle AED=\angle EAB$$
This implies $OE$ is tangent to circumcircle of $\triangle ABE$. But $ABE$ is a right triangle with its circumcenter on $AE$. So $OE \perp AE$ which means $AE$ is tangent to the Apollonian circle $(O)$.