Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$ Question:

Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$.

My try:
$$\begin{align}\\
&x^2+y^2-4x=0\\
&\implies (x-2)^2+y^2=4\\
&\implies |z-2|=2\\
\end{align}\\
$$
Now, $w=\frac{2z+3}{z-4}$
$$\begin{align}\\
&\frac w2=\frac{2z+3}{2z-8}\\
&\implies\require{cancel}\frac{w}{w-2}=\frac{2z+3}{\cancel{2z}+3-\cancel{2z}+8}\\
&\implies\frac{w}{w-2}=\frac{2z+3}{11}\\
&\implies\frac{2z}{11}=\frac{w}{w-2}-\frac{3}{11}\\
&\implies\frac{2z}{\cancel{11}}=\frac{8w+6}{\cancel{11}(w-2)}\\
&\implies z=\frac{4w+3}{w-2}\\
&\implies z-2=\frac{2w+7}{w-2}\\
\end{align}\\
$$
$$\therefore\left|\frac{2w+7}{w-2}\right|=2\\
\implies 2w+7=2w-4
$$
Now, what to do? Where is my fault? Or how to do it? Is there any other possible ways?

Ak19 answered what I would have answered before I got to it, so here's another possible way of how to do it.  These transformations map generalized circles to generalized circles; a generalized circle is either a circle or a line.  This particular transformation maps the point $z=0$ on the given circle to $w=-\frac34$, the point $z=2+2i$ on the given circle to $-\frac34-\frac{11}4i$, and the point $z=2-2i$ on the given circle to $-\frac34+\frac{11}4i$, so we can see that it maps the circle to the line $u=-\frac34$.
