Equation of Circle in Complex Plane/Formulation 1

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 $1$.