Show that general bilinear transformation transforms circles into circles (or lines) The exercise I got a bit stuck with is at the end of chapter on complex numbers that does not deal with transformations in general. So the notion of general bilinear transformation is introduced for the purpose of this exercise on the spot. 
I am mentioning this because the answer I am looking for should, therefore, be made by more or less brute-force caluculation, not by using general results concerning transformations of complex numbers. 
What I should show is that general bilinear transformation, defined as transformation from $z$-plane to $Z$-plane via 
$$z = \frac{a Z + b}{c Z + d}$$ 
transforms a circle given by 
$$\left|\frac{z-z_1}{z-z_2}\right| = H$$ 
where $H$ is a real number not equal to $1$, into a circle or straight line, and then determine what conditions $z_1$, $z_2$ and $H$ should satisfy in order for the circle to be transformed into a straight line.
Now, when I substitute $(aZ+b)/(cZ+d)$ for $z$ in the given equation of the circle I get
$$\left|\frac{(a - c z_1)Z + b - d z_1}{(a - c z_2 ) Z + b - d z_2}\right| = H$$
and from here it seems immediately clear to me that when I raise both sides to the power of 2 and simplify the equation I will get a general conic section curve 
$Ax^2 + By^2 + Cx + Dy + \cdots =0$ 
in which $A$ won't be equal to $B$ (due simply to the difference between $z_1$ and $z_2$) and so it won't be an equation of the circle. 
Where am I making a mistake? 

Converting comment to answer, as requested ...

When you square and simplify, you'll get (in part) ...
$$\left| a - c z_1 \right|^2 \left|Z\right|^2 + \cdots = H^2 \left( \left|a-c z_2\right|^2 \left|Z\right|^2 + \cdots \right)$$
so that 
$$ 0 = \left|Z\right|^2\;\left( \left|a-c z_1\right|^2 - H^2 \left|a - c z_2\right|^2 \right) + \cdots = \left( x^2 + y^2 \right) A + \cdots$$
The "$\cdots$" stuff involves single powers of $Z$ and $\overline{Z}$ (and thus $x$ and $y$), and constants. The coefficients of $x^2$ and $y^2$ definitely match.
