Intuitively, why should I expect a circle in the complex plane from the equation $\left|\frac{z-1}{z+1}\right| = c$?
Going in steps:
- $|\frac12 - z| = \frac c2$, or $|1 - 2z| = c$, is the intuitive representation of a circle.
- $|1 - 2\overline{z}| = c$ is still a circle; we've just reflected it about the real axis.
- Here's the tricky part. Replacing $\overline{z}$ by $\frac1z$ preserves the angle of a point but takes the reciprocal of the magnitude. This makes it an inversion in the complex plane, which also preserves circles. (Well, circles through the origin become lines instead, but our circle isn't one of those.) So $|1 - \frac2z| = c$ is still a circle.
- Finally, going from $|\frac{z-2}{z}| = c$ to $|\frac{z-1}{z+1}| = c$ is just a translation by $1$.
Let's explore. $$\bbox{ \left\lvert \frac{z - 1}{z + 1} \right\rvert = c } \quad \iff \quad \bbox{ \frac{\lvert z - 1 \rvert}{\lvert z + 1 \rvert} = c } \quad \iff \quad \bbox{ \left\lvert z - 1 \right\rvert = c \left\lvert z + 1 \right\rvert } \tag{1}\label{NA1}$$ This only makes sense if $0 \lt c \in \mathbb{R}$. Since $z \in \mathbb{C}$, we can write $z = x + i y$. Since $\lvert z \rvert = \sqrt{x^2 + y^2}$, we have $$\bbox{ \left\lvert \frac{z - 1}{z + 1} \right\rvert = c } \quad \iff \quad \bbox{ \sqrt{(x-1)^2 + y^2} = c \sqrt{(x+1)^2 + y^2} } \tag{2}\label{NA2}$$ Since $x, y, z \in \mathbb{R}$ and $c \gt 0$, both sides are positive, and we can square both sides: $$\bbox{ \left\lvert \frac{z - 1}{z + 1} \right\rvert = c } \quad \iff \quad \bbox{ (x-1)^2 + y^2 = c^2 (x+1)^2 + c^2 y^2 } \tag{3}\label{NA3}$$ Expanding and moving all terms to one side, we get $$\bbox{ \left\lvert \frac{z - 1}{z + 1} \right\rvert = c } \quad \iff \quad \bbox{ x^2 - 2 x + 1 + y^2 - c^2 x^2 - c^2 y^2 - 2 c^2 x - c^2 = 0 } \tag{4}\label{NA4}$$ The case when $c = 1$ is special, because then $\eqref{NA3}$ simplifies to $x = 0$, which is not a circle but a line (unless you say it is an infinite-radius circle centered at real $\pm\infty$). In any case, let's continue the exploration with $0 \lt c \in \mathbb{R}$, $c \ne 1$.
We can collect the terms in $\eqref{NA4}$, getting $$\bbox{ \left\lvert \frac{z - 1}{z + 1} \right\rvert = c } \quad \iff \quad \bbox{ (1 - c^2)\left( (x + 1)^2 - \frac{4 x}{1 - c^2} + y^2 \right) = 0 } \tag{5}\label{NA5}$$ Because we already decided $c \ne 1$, this is equivalent to $$\bbox{ \left\lvert \frac{z - 1}{z + 1} \right\rvert = c } \quad \iff \quad \bbox{ (x + 1)^2 - \frac{4 x}{1 - c^2} + y^2 = 0 , \quad c \ne 1 } \tag{6}\label{NA6}$$ This is getting interesting. Compare to the equation of a circle of radius $r$ centered at $x = x_0$, $$\bbox{ (x - x_0)^2 + y^2 - r^2 = 0 }$$ Now, if we choose $$\bbox{ x_0 = \frac{2}{1 - c^2} - 1} , \quad \bbox{ r = \sqrt{\left(\frac{1 + c^2}{1 - c^2} \right)^2 - 1} }$$ we find that $$\bbox{ (x - x_0)^2 + y^2 - r^2 = (x + 1)^2 - \frac{4 x}{1 - c^2} + y^2 }$$ Therefore, $$\bbox[#ffffef]{ \bbox{ \left\lvert \frac{z - 1}{z + 1} \right\rvert = c } \quad \iff \quad \bbox{ (1 - c^2)\left( (x - x_0)^2 + y^2 - r^2 \right) = 0 } , \quad \bbox{ z = x + i y } \tag{7a}\label{NA7a} }$$ where $$\bbox[#ffffef]{ \bbox{ x_0 = \frac{2}{1 - c^2} - 1 } , \quad \bbox{ r = \sqrt{\left(\frac{1 + c^2}{1 - c^2} \right)^2 - 1} } , \quad \bbox{ c \gt 0 } , \quad \bbox{ c \ne 1 } , \quad \bbox{ c \in \mathbb{R} } \tag{7b}\label{NA7b} }$$ and describes a circle of radius $r$ centered at $z = x_0$ on the real axis when $c \gt 0$, $c \ne 1$, and a line along the imaginary axis when $c = 1$. No intuition or geometry needed, basic algebra suffices.
Hint:
$$ \eqalign{ & \left| {{{z - 1} \over {z + 1}}} \right| = c\quad \Leftrightarrow \quad \left| {z - 1} \right| = c\left| {z + 1} \right|\quad \Leftrightarrow \cr & \Leftrightarrow \quad {\rm distance}\;\left( {x,y} \right)\;{\rm from}\;(1,0) = c\; \cdot \;{\rm distance}\;\left( {x,y} \right)\;{\rm from}\;( - 1,0) \cr} $$ which is another way to define a circle.
Thanks to @Rahul for indicating the actual attribution for such definition ( https://en.wikipedia.org/wiki/Circle#Circle_of_Apollonius )