Equation of Circle in Complex Plane/Formulation 2/Proof 2

Proof
From Equation of Circle in Cartesian Plane: Corollary 1, the equation for a circle is:


 * $A \left({x^2 + y^2}\right) + B x + C y + D = 0$

provided that:
 * $B^2 + C^2 \ge 4 A D$
 * $A > 0$.

Thus:

By setting:
 * $\alpha := A$, $\beta := \dfrac B 2 + \dfrac C {2 i}$ and $\gamma := D$

we have:
 * $\alpha z \overline z + \beta z + \overline \beta \overline z + \gamma = 0$

As $A, B, C, D \in \R$ it follows that both $\alpha$ and $\gamma$ are real.

Then we have that $A > 0$ and so $\alpha > 0$.

Then note that:

Given that:
 * $B^2 + C^2 \ge 4 A D$

it follows that:
 * $\cmod \beta^2 > \alpha \gamma$

If $\alpha = $ and $\beta \ne 0$ the equation devolves to:


 * $\beta z + \overline \beta \overline z + \gamma = 0$

which from Equation of Line in Complex Plane: Formulation 1 is the equation of a straight line.

The result follows.