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?

This is equivalent to showing that $4w+3$ is pure imaginary i.e. $\arg(4w+3)=\pm\frac{\pi}{2}$.
Since $0$ and $4$ are end points of the circle diameter and $4w+3=11\frac{z-0}{z-4}$, then $\arg(4w+3)=\pm\frac{\pi}{2}$.
