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?

You can't replace modulus like that in the last line ( because complex numbers with equal magnitude may not be actually equal. For e.g., $2i$ and $-2i$ have magnitude $2$ but aren't equal.)
Use $w = u+iv$
So, $|2(u+iv)+7| = 2|(u+iv)-2|$
Evaluating and squaring both sides,
$\Rightarrow (2u+7)^2 + 4v^2 = 4(u-2)^2 + 4v^2 $
$\Rightarrow 4u^2 + 28u +49  = 4u^2 - 16u + 16 $
$\Rightarrow 44u + 33 = 0 $
$\Rightarrow 4u +3 =0 $
