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 $$\frac{k}{1-k^2} |a-b|$$