Apollonius Circle

Locus of z such that its distances from z₁ and z₂ are in constant ratio k. For k ≠ 1, this is a circle (Apollonius circle). For k = 1, it degenerates to the perpendicular bisector of z₁z₂. The centre and radius can be found by squaring and expanding.

The locus $|z - z_1|/|z - z_2| = k$ looks like a ratio condition. For $k \neq 1$, it is secretly a circle — a fact not obvious until you expand it.

Squaring the Condition

$$\left|\frac{z - z_1}{z - z_2}\right| = k \implies |z - z_1|^2 = k^2|z - z_2|^2$$

Expanding each side using $|w|^2 = w\bar{w}$:

$$(z - z_1)(\bar{z} - \bar{z}_1) = k^2(z - z_2)(\bar{z} - \bar{z}_2)$$ $$z\bar{z} - z_1\bar{z} - \bar{z}_1 z + |z_1|^2 = k^2\left(z\bar{z} - z_2\bar{z} - \bar{z}_2 z + |z_2|^2\right)$$

Rearranging

Bring all terms to the left:

$$(1 - k^2)z\bar{z} + (k^2 z_2 - z_1)\bar{z} + (\overline{k^2 z_2 - z_1})z + |z_1|^2 - k^2|z_2|^2 = 0$$

For $k \neq 1$, divide through by $(1 - k^2)$:

$$z\bar{z} + \frac{k^2 z_2 - z_1}{1 - k^2}\bar{z} + \overline{\left(\frac{k^2 z_2 - z_1}{1 - k^2}\right)}z + \frac{|z_1|^2 - k^2|z_2|^2}{1 - k^2} = 0$$

This is the general circle form $z\bar{z} + \bar{a}z + a\bar{z} + b = 0$ with:

$$a = \frac{k^2 z_2 - z_1}{1 - k^2}, \qquad b = \frac{|z_1|^2 - k^2|z_2|^2}{1 - k^2} \in \mathbb{R}$$

Centre and Radius

Centre: $-a = \dfrac{z_1 - k^2 z_2}{1 - k^2}$

Radius: $\sqrt{|a|^2 - b}$ — computable from the values above.

The Special Case $k = 1$

When $k = 1$, the factor $(1 - k^2) = 0$ and the $z\bar{z}$ terms cancel. What remains is:

$$(z_2 - z_1)\bar{z} + \overline{(z_2 - z_1)}z + |z_1|^2 - |z_2|^2 = 0$$

This is linear in $x$ and $y$ — a straight line, specifically the perpendicular bisector of $z_1 z_2$. So the family of Apollonius loci transitions continuously from circles ($k \neq 1$) to the perpendicular bisector ($k = 1$) as $k \to 1$.