How to prove that a Möbius image is a circle? Let $\mu(z) = \frac{z-ia}{z-ib}$, $z=x+iy$, where $x,y, a, b$ are real, with $a<b$. Define a line $l = x+ia$. We can see that the image of this line under $\mu$ is not a line:
$$\mu(l)=\frac{x}{x+i(a-b)}.$$
We also know that Möbius transformations map lines to lines or circles in $\mathbb{C}$. But the problem is - I can see that $\mu(l)$ is not a line, but can't see how it is a circle.
Here's what I do:
-> Take the modulus^2 of the equation above to get
$$ \mu^2 = \frac{x^2}{ x^2+(a-b)^2 }.$$
-> After some algebra I get
$$ (x\mu)^2 + (a-b)^2\mu^2 - x^2 =0 $$
But is this a circle? Actually, WolframAlpha doesn't think so. What is it that I'm not doing right?

The circle is formed by $\Re(\mu(x))$ and $\Im(\mu(x))$, with $x$ acting as a parameter. To avoid confusion I will call the parameter $t$, and use $c = a - b$.
Obtain real and imaginary parts as $$\frac{t}{t - ic} = \frac{t^2 + ict}{t^2 + c^2} = \frac{t^2}{t^2+c^2} + i\frac{ct}{t^2+c^2}$$ Now let $x = \frac{t^2}{t^2+c^2}$ and $y = \frac{ct}{t^2+c^2}$. Exclude $t$ to obtain and equation on $x, y$:
$xt^2 = t^2 + c^2$, so $t^2 = \frac{c^2}{1 - x}$ (and $t^2 + c^2 = xt^2 = \frac{c^2x}{1 - x}$).
Therefore, $y^2 = \frac{c^2t^2}{(t^2 + c^2)^2} = c^2t^2 (\frac{1-x}{c^2})^2 = c^2 \frac{c^2}{1-x} \frac{(1-x)^2}{c^4} = x(1-x)$, or
$$(x - \frac{1}{2})^2 + y^2 = \frac{1}{4}$$
