Trouble in mapping of möbius transformation 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 attempt:-
The circle $x^2+y^2-4x=0$ is $|z-2|=2$ . . .$(1)$
So the inverse mapping of the given bilinear transformation is:-
$$z= \frac{4w+3}{w-2} $$
Now substituting  the value of $z$ in  $(1)$
$$\frac{|3w+1|}{|w-2|} =2$$
$$|4w+3|=2|w-2|$$
$$|3u+2+3v\iota|=2|u-2+v\iota|$$
$$9u^2+4+12u+v^2= 4u^2+16-16u+v^2$$
On solving these it appears as
$$5u^2+28u-12=0$$
I can not come at the conclusion as stated in question, is my method correct ?
Suggestions are highly appreciated 
Thankyou

If $z= x+yi$ then equation of circle is $$|z-2|=2$$ Since $z= (4w+3)/(w-2)$ we get
$$\Big|{4w+3 -2w+4\over w-2}\Big| = 2$$  or $$|2w+7|= |2w-4|$$ dividing this by $2$ we get: $$|w-(-7/2)|=|w-2|$$
so $w$ is on perpendicular bisector between $-7/2$ and $2$ so $w=-{3\over 4}$ or $$4w+3=0$$
