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

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Theorem

Let $\C$ be the complex plane.

Let $C$ be a circle in $\C$.


Then $C$ may be written as:

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

where:

$\alpha \in \R_{\ne 0}$ is real and non-zero
$\gamma \in \R$ is real
$\beta \in \C$ is complex such that $\cmod \beta^2 > \alpha \gamma$.

The curve $C$ is a straight line if and only if $\alpha = 0$ and $\beta \ne 0$.


Proof

From Equation of Circle in Cartesian Plane: Formulation 2, the equation for a circle is:

$A \paren {x^2 + y^2} + B x + C y + D = 0$

provided that:

$B^2 + C^2 \ge 4 A D$
$A > 0$.

Thus:

\(\ds A \paren {x^2 + y^2} + B x + C y + D\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds A z \overline z + B x + C y + D\) \(=\) \(\ds 0\) Product of Complex Number with Conjugate
\(\ds \leadsto \ \ \) \(\ds A z \overline z + \frac B 2 \paren {z + \overline z} + C y + D\) \(=\) \(\ds 0\) Sum of Complex Number with Conjugate
\(\ds \leadsto \ \ \) \(\ds A z \overline z + \frac B 2 \paren {z + \overline z} + \frac C {2 i} \paren {z - \overline z} + D\) \(=\) \(\ds 0\) Difference of Complex Number with Conjugate
\(\ds \leadsto \ \ \) \(\ds A z \overline z + \paren {\frac B 2 + \frac C {2 i} } z + \paren {\frac B 2 - \frac C {2 i} } \overline z + D\) \(=\) \(\ds 0\) gathering terms

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:

\(\ds \paren {\frac B 2 + \frac C {2 i} } \paren {\frac B 2 - \frac C {2 i} }\) \(=\) \(\ds \paren {\frac B 2}^2 + \paren {\frac C 2}^2\)
\(\ds \) \(=\) \(\ds \frac {B^2 + C^2} 4\)

Given that:

$B^2 + C^2 \ge 4 A D$

it follows that:

$\cmod \beta^2 > \alpha \gamma$

$\Box$


If $\alpha = 0$ 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.

$\blacksquare$


Sources

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