Apollonius circle, its radius and center

The algebra is messy but straightforward. Let $z=x+i y$, $a=a_r+i a_i$ and $b=b_r+i b_i$. Square both sides of the defining equation to get

$$(x-a_r)^2+(y-a_i)^2 = k^2 (x-b_r)^2 + k^2 (y-b_i)^2$$

Rearrange and expand to get

$$(1-k^2) x^2 + (1-k^2) y^2 - 2 (a_r-k^2 b_r) x - 2 (a_i - k^2 b_r) y + |a|^2-k^2 |b|^2 = 0$$

Now complete the square. This is where it gets messy.

$$\begin{align}(1-k^2) \left ( x-\frac{a_r-k^2 b_r}{1-k^2}\right )^2 + (1-k^2) \left ( y-\frac{a_i-k^2 b_i}{1-k^2}\right )^2 \\= \frac{(a_r-k^2 b_r)^2+(a_i-k^2 b_i)^2}{1-k^2} - (|a|^2-k^2 |b|^2)\\ = \frac{|a|^2 + k^4 |b|^2 - 2 k^2 (a_r b_r+a_i b_i) - (|a|^2-k^2 |b|^2)(1-k^2)}{1-k^2}\\ = \frac{k^2 (|a|^2+|b|^2) - 2 k^2 (a_r b_r+a_i b_i)}{1-k^2}\\ = \frac{k^2}{1-k^2} |a-b|^2\end{align}$$

From here, I hope it is clear that the above reduces to, in complex notation:

$$\left |z-\frac{a-k^2 b}{1-k^2}\right| = \frac{k}{1-k^2} |a-b|$$

So, a circle of center $$\frac{a-k^2 b}{1-k^2}$$ and radius $$\frac{k}{1-k^2} |a-b|$$

This is an old post, but I wanted to see if the calculation could be illuminated by unpacking to real coordinates later, and by solving a simpler equation. After doing this I show how to do the calculation without unpacking to real coordinates.

We want to show that the set of $z$ satisfying $|z - a| = k |z - b|$ forms a circle.

We can simplify our calculation by noting that translation preserves circles. We can translate to $w = z - b$ and solve the equation $|w - A| = k|w|$, where $A = a - b$, and then translate our solution back to $z$.

Note that $$|w-A|^2 = (w-A) (\overline{w-A}) = |w|^2 - 2 \mathrm{Re}(\bar{A}w) + |A|^2.$$ This is a version of the polarization identity, noting that $\mathrm{Re}(\bar{A} z)$ is the Euclidean dot product between $A$ and $z$.

Then we see

\begin{align*} |w - A| = k |w| &\Longrightarrow |w - A|^2 = k^2 |w|^2 \\ &\Longrightarrow |w|^2 - 2 \mathrm{Re}(\bar{A}w) + |A|^2 = k^2 |w|^2 \\ &\Longrightarrow (1 - k^2)|w|^2 - 2\mathrm{Re}(\bar{A} w) + |A|^2 = 0 \ \ \ (\star) \end{align*}

Let $w = x + i y$ and $A = A_x + i A_y$. If $k \neq 1$, then \begin{align*} (\star) &\Longrightarrow |w|^2 - \frac{2}{1 - k^2} \mathrm{Re}(\bar{A}w) + \frac{|A|^2}{1 - k^2} = 0 \\ &\Longrightarrow x^2 + y^2 - \frac{2 A_x}{1 - k^2}x - \frac{2 A_y}{1 - k^2}y + \frac{|A|^2}{1 - k^2} = 0 \end{align*}

Completing the square shows that sets determined by an equation in the following form are always (possibly degenerate) circles. \begin{align*} x^2 + y^2 + Ex + Fy + G = 0 \Rightarrow (x + \frac{E}{2})^2 + (y + \frac{F}{2})^2 = \frac{E^2 + F^2}{4} - G. \end{align*}

Replacing coefficients in this formula with those of our previous expression, we find \begin{align*} \left(x - \frac{A_x}{1 - k^2} \right)^2 + \left(y - \frac{A_y}{1 - k^2}\right)^2 = \frac{A_x^2 + A_y^2}{(1 - k^2)^2} - \frac{|A|^2}{1 - k^2} = \left( \frac{k |A|}{|1 - k^2|} \right)^2 \end{align*}

Since $A = a - b$, we confirm that the radius is $r = k\frac{|a - b|}{|1 - k^2|}$. In complex form, we find $$|w - \frac{a - b}{1 - k^2}| = \frac{k|a - b|}{|1 - k^2|},$$ Translating back to $z$, we find $$|z - \frac{a - b k^2}{1 - k^2}| = \frac{k|a - b|}{|1 - k^2|},$$ which shows that the center is $c = \frac{a - b k^2}{1 - k^2}$.

There's a slicker calculation which doesn't unpack to real coordinates. It uses the polarization identity again, which serves as a complex substitute for completing the square: $$|z|^2 - 2 \mathrm{Re}(\bar{d}z) = |z - d|^2 - |d|^2$$

From $(\star)$, \begin{align*} 0 &= |w|^2 - 2\mathrm{Re}(\frac{\bar{A}}{1 - k^2}w) + \frac{|A|^2}{1 - k^2} \\ &= | w - \frac{A}{1 - k^2} |^2 - |\frac{A}{1 - k^2}|^2 + \frac{|A|^2}{1 - k^2}, \end{align*} which implies \begin{align*} | w - \frac{A}{1 - k^2} |^2 = \left( \frac{k |A|}{1 - k^2} \right)^2. \end{align*} Then we can finish as before.