# Circle of Apollonius in Complex Plane

Jump to navigation
Jump to search

## Theorem

Let $\C$ be the complex plane.

Let $\lambda \in \R$ be a real number such that $\lambda \ne 0$ and $\lambda \ne 1$.

Let $a, b \in \C$ such that $a \ne b$.

The equation:

- $\cmod {\dfrac {z - a} {z - b} } = \lambda$

decribes a circle of Apollonius $C$ in $\C$ such that:

- if $\lambda < 0$, then $a$ is inside $C$ and $b$ is outside
- if $\lambda > 0$, then $b$ is inside $C$ and $a$ is outside.

If $\lambda = 1$ then $z$ describes the perpendicular bisector of the line segment joining $a$ to $b$.

## Proof

By the geometry, the locus described by this equation is a circle of Apollonius.

This needs considerable tedious hard slog to complete it.In particular: etc.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Examples

### Example: $\cmod {\dfrac {z - 3} {z + 3} } = 2$

The equation:

- $\cmod {\dfrac {z - 3} {z + 3} } = 2$

describes a circle embedded in the complex plane whose radius is $4$ and whose center is $\paren {-5, 0}$.

### Example: $\cmod {\dfrac {z - 3} {z + 3} } < 2$

The inequality:

- $\cmod {\dfrac {z - 3} {z + 3} } < 2$

describes the exterior of a circle embedded in the complex plane whose radius is $4$ and whose center is $\paren {-5, 0}$.

## Sources

- 1960: Walter Ledermann:
*Complex Numbers*... (previous) ... (next): $\S 2$. Geometrical Representations: Exercise $8$