Equation of Circle in Complex Plane/Formulation 1
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Theorem
Let $\C$ be the complex plane.
Let $C$ be a circle in $\C$ whose radius is $r \in \R_{>0}$ and whose center is $\alpha \in \C$.
Then $C$ may be written as:
- $\cmod {z - \alpha} = r$
where $\cmod {\, \cdot \,}$ denotes complex modulus.
Interior
The points in $\C$ which correspond to the interior of $C$ can be defined by:
- $\cmod {z - \alpha} < r$
Exterior
The points in $\C$ which correspond to the exterior of $C$ can be defined by:
- $\left\lvert{z - \alpha}\right\rvert > r$
Proof
Let $z = x + i y$.
Let $\alpha = a + i b$.
Thus:
\(\ds \cmod {z - \alpha}\) | \(=\) | \(\ds r\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \cmod {x + i y - a + i b}\) | \(=\) | \(\ds r\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \cmod {\paren {x - a} + i \paren {y - b} }\) | \(=\) | \(\ds r\) | Definition of Complex Subtraction | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sqrt {\paren {x - a}^2 + \paren {y - b}^2}\) | \(=\) | \(\ds r\) | Definition of Complex Modulus | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x - a}^2 + \paren {y - b}^2\) | \(=\) | \(\ds r^2\) | squaring both sides |
The result follows from Equation of Circle.
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations: Example $1$.