Although $(R,h)$ are variable, difference of their squared distances $ (R^2-h^2)= $ should be constant $a^2$ as shown in the picture following.
So with every fixed point A1 there is a corresponding fixed point A2 mirrored about $x-$ axis.
$$ (x-h)^2+ y^2 = h^2 + a^2 \,or \quad x^2-2xh +y^2 = a^2 \tag1 $$
where $h$ is variable point $B$ is moving on x-axis and $a$ is constant. The above circle with $B$ as center is now drawn in black.
EDIT1:
Points $(A1,A2),(OA1=OA2=a)$ are singular points, meaning all circles pass through them. They are obtained as singular solutions of DE derived by partial differentiation with respect
to $h$ and eliminating it from 1)
$$ x^2-2 x\,h +y^2=a^2\, \rightarrow x =0 \rightarrow y= \pm a \tag2 $$
The distance $OA$ in fact serves as one among two bipolar $\sigma$ isosurface /coordinates.
EDIT2
Please note that, if you had asked:
Prove that all the circles having centres on a fixed line and whose tangents pass through a fixed point , lying on that fixed line must have another tangent passing through this fixed point,
then the constant in this case gets its sign simply changed to $ (h^2-R^2)= a^2 $ when we are now switching to their orthogonal trajectories:
$\tau $ isosurface /coordinates
which have equi-tangent constant lengths $=a$ circles of equation
$$ x^2+y^2-2 x h -R^2= a^2 \tag3 $$
I have included this as your preference is for the analytical approach. But do not lose the geometrical connection!
To make the question answered, here is the solution; each such line passes from one vertex of the Johnson triangle to the other homothetic counterpart in the reference triangle. The homothetic center should be on each line, hence the lines are concurrent.
Best Answer
Let the circle be $|z-b|=r$. Then $$|a-b|^2 = |1/\bar a - b|^2 = r^2.$$ Expand to get $$ |a|^2 - 2 \Re(\bar ab) + |b|^2 = r2,\qquad 1/|a|^2 - 2\Re(b/a) + |b|^2 = r^2.$$
Compare to get $$ 2\Re[(|a|^2-1)b/a] = 2\Re(\bar ab - b/a) = |a|^2 - 1/|a|^2.$$
It is clear that $|a|\neq 1$; otherwise $a = 1/\bar a$. So $$ 2\Re(b/a) = 1+ 1/|a|^2.$$
Substitute back to see $$|b|^2 = r^2+1.$$
On the other hand, let $z_0$ be an intersection of the two circles: $$ |z_0-b|=r,\qquad |z_0|=1.$$ Expand to get $$ r^2 = |z_0|^2 - 2\Re(\bar z_0b) + |b|^2 = 1 -2\Re(b/z_0) + |b|^2.$$ It follows that $$ \Re[(z_0-b)/z_0] = 1-\Re(b/z_0) = 1 + (r^1-1-|b|^2)/2 = 0.$$ The radii from $z_0$ to $0$ and from $z_0$ to $b$ are orthogonal. So the two circles intersect at right angles.