I am having a problem due to what no doubt must be an embarrassing error on my part.
I have a Moebius transformation $\phi(z)=(a z+b)/(c z+d)$ with $a,b,c,d$ complex. As $t$ goes over the reals, $\phi(t)$ moves in a circle. Assume $a$ and $c$ both equal to a complex number $w$ of modulus $\sqrt{x}$ ($x$ real), $b = -(1+x)$ and d = $-(1+w^2)$. According to Ch II, section 41 in Carathéodory's book, the circle has center $$\frac{a \overline{d}- b \overline{c}}{c \overline{d} - d\overline{c}}$$. Here the denominator is $$- w (1 + \overline{w}^2) + \overline{w} (1 + w^2) = - w - \overline{w} x + \overline{w} + w x = (\overline{w} - w) (1 - x).$$ The numerator equals $$- w (1 + \overline{w}^2) + (1+x) \overline{w} = \overline{w} - w$$ (since $x = w \overline{w}$). Hence, the center has modulus $1/(1-x)$. However, it is clear that, as w approaches the real number $\sqrt{x}$ (on the circle of radius $\sqrt{x}$ around the origin), it is clear that our circle "tends" towards the circle with center at $1$ and radius $0$. What gives?
Oh, and the radius is supposed to be $$\frac{|a d - b c|}{|c \overline{d} - \overline{c} d|},$$ which is $$\frac{|-w (1 + w^2) + (1+x) w|}{|(\overline{w}-w) (1-x)|} = \frac{|x w - w^3|}{|(\overline{w}-w) (1-x)|} = \frac{|w^2|}{|1-x|} = \frac{x}{|1-x|},$$ which, again, does not approach $0$ as $w$ approaches $\sqrt{x}$. Again: what gives?
(At least the circle goes through a point at distance $1$ from the origin, reflecting the fact that $ \phi(t)\to 1$ as $t\to \pm \infty$. But this is not the circle we want -- it may be the circle we would have if we let $w\to i \sqrt{x}$.)
Another possible approach to try:
Invert the equation $w= \phi(z)$ as $z= \psi(w)$. Determine the inverse explicitly in terms of $a,b,c,d$. (You can find the inverse of the matrix associated to the original linear-fractional map.)
Then write the real-axis condition $ z -\overline z =0$ as an algebraic equation in the symbols $w$ and $\overline w$. Manipulate that to get the complex-variable equation of your circle: $(w-c) \overline {(w-c)}= r^2$.