Let $\alpha,r\in\mathbb{R}$ with $r>0$ and $|\alpha|+r\le 1$, and consider the fractional linear transform $$ f(z) = \frac{z-\alpha}{1-\alpha z}. $$ I would like to show the following: the circle in $C(\alpha,r)\subset\mathbb{C}$ centered at $\alpha$ with radius $r$ is mapped onto a circle of larger radius by $f$.
My approach: Note that $f$ maps $[-1,1]$ into $[-1,1]$ with $f(-1)=-1$ and $f(1)=1$, and in particular it maps the line segment $[\alpha-r,\alpha+r]$ to some line segment $[f(\alpha-r),f(\alpha+r)]$ on the real line. Furthermore, the line segment $[\alpha-r,\alpha+r]$ is perpendicular to the boundary of $C(\alpha,r)$ at the intersection points $\alpha\pm r$, and since conformal maps preserve angles, the image of the line segment is also perpendicular to the boundary of $f(C(\alpha,r))$ at $f(\alpha\pm r)$. Thus, $[f(\alpha-r),f(\alpha+r)]$ is a diameter of the image of $f(C(\alpha,r))$, and so if $r'$ is the radius of the $f(C(\alpha,r))$, then $2r' = f(\alpha+r)-f(\alpha-r)$. Thus, we have \begin{align} r' = \frac{1}{2}(f(\alpha+r)-f(\alpha-r)) &= \frac{1}{2}\left(\frac{\alpha+r-\alpha}{1-\alpha(\alpha+r)}-\frac{\alpha-r-\alpha}{1-\alpha(\alpha-r)}\right) \\ &=\frac{r}{2}\left(\frac{1}{1-\alpha^2-\alpha r}+\frac{1}{1-\alpha^2+\alpha r}\right) \\ &=r\left(\frac{(1-\alpha^2)}{(1-\alpha^2)^2-(\alpha r)^2}\right) \end{align} and hence $$\frac{r}{r'} = \frac{(1-\alpha^2)^2-(\alpha r)^2}{1-\alpha^2}\le\frac{(1-\alpha^2)^2}{1-\alpha^2}\le 1\implies r'\ge r.$$
I would like to know if there is a simpler approach to show that the radius increases (since right now it appears just to be algebra magic), or if there is a non-rigorous but intuitive explanation for this effect. In particular, I'm wondering if this has anything to do with hyperbolic distances, given that hyperbolic distances are larger with respect to Euclidean ones near the boundary of the unit disk and smaller near the center.
We can happily decompose $f$ as follows:$$f(z) = {{z - \alpha}\over{1 - \alpha z}} = {{-1}\over\alpha} + {{1 - {1\over{\alpha^2}}}\over{z - {1\over\alpha}}}.$$If$$g(z) = z - {1\over\alpha}, \quad h(z) = \left(1 - {1\over{\alpha^2}}\right)z, \quad i(z) = {1\over z},$$we have$$f(z) = g\left({{1 - {1\over{\alpha^2}}}\over{z - {1\over\alpha}}}\right) = g \circ h\left({1\over{z - {1\over\alpha}}}\right) = g \circ h \circ i \circ g(z).$$$g$ has no effect on radii, $h$ which change radii by a factor of $|1 - 1/\alpha^2|$. But the effect of $i$ on radii is essentially working through the above approach of calculating $(1/2)(f(\alpha + r) - f(\alpha - r))$, just with $f$ replaced with $i \circ g$. So to be honest, I do not think the above "algebra magic" is in any way magic, but more the actual reason as to the truth.
The question itself is not particularly deep, so it is not hugely surprising to me that the solution did not require anything deep.