Equation of Circle in Complex Plane/Formulation 2
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 1
\(\ds \cmod {z - a}\) | \(=\) | \(\ds r\) | Equation of Circle in Complex Plane: Formulation 1 | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \cmod {z - a}^2\) | \(=\) | \(\ds r^2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {z - a} \overline {\paren {z - a} }\) | \(=\) | \(\ds r^2\) | Modulus in Terms of Conjugate | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {z - a} {\paren {\overline z - \overline a} }\) | \(=\) | \(\ds r^2\) | Sum of Complex Conjugates | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds z \overline z - a z - \overline a \overline z + a \overline a\) | \(=\) | \(\ds r^2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha z \overline z - \alpha a z - \alpha \overline a \overline z + \alpha a \overline a - r^2\) | \(=\) | \(\ds 0\) | where $\alpha \in \R_{>0}$ is arbitrary |
By setting:
- $\beta := -\alpha a$ and $\gamma = \alpha a \overline a - r^2$
we have:
- $\alpha z \overline z + \beta z + \overline \beta \overline z + \gamma = 0$
We have that:
\(\ds r^2\) | \(>\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha a \overline a - \gamma\) | \(>\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\alpha a} \paren {\alpha \overline a}\) | \(>\) | \(\ds \alpha \gamma\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \beta \overline \beta\) | \(>\) | \(\ds \alpha \gamma\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \cmod {\beta}^2\) | \(>\) | \(\ds \alpha \gamma\) | Modulus in Terms of Conjugate |
$\Box$
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.
$\blacksquare$
Proof 2
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.
$\blacksquare$