Proving the Apollonian circle formula

Proposition: Let: $c,d\in\mathbb{C} \space : c\neq d$ Then: $\forall k \in \mathbb{R}^{+}\setminus\{1\} \forall z \in \mathbb{C}: |z-c|=k|z-d|$ represents Appolonius circle.

What was already done: Considering $\{z\in\mathbb{C}: Az\bar{z}+Bz+\bar{Bz}+C=0, A,C\in\mathbb{R}, B\in \mathbb{C} \}$ this set represents circle when $\begin{cases}|B|^{2}>AC\\ A\in\mathbb{R}\setminus\{0\}\\ C\in\mathbb{R}\\ \end{cases}$

Have intuition that it is sufficient to derive formula for the centre and radius of the circle. However, I am not sure whether it is enough to prove such statement, especially to obtain all properties of Appolonius circle.

Request Will be very thankful for hints/helps!!!

Hint (skipping over some intermediate steps):

$$\require{cancel} \begin{align} 0 = |z-c|^2 - k^2|z-d|^2 &= (z-c)(\bar z - \bar c) - k^2(z-d)(\bar z - \bar d) \\ &= z \bar z (1-k^2)- z(\bar c - k^2 \bar d) - \bar z(c- k^2 d)+ c \bar c - k^2 d \bar d \\ &= (1-k^2)\left(z - \frac{c - k^2 d}{1-k^2}\right)\left(\bar z - \frac{\bar c - k^2 \bar d}{1-k^2}\right) - \frac{(c-k^2d)(\bar c - k^2\bar d)}{1-k^2}+c\bar c-k^2 d\bar d \\ &= (1-k^2)\,\left|z -\frac{c-k^2d}{1-k^2}\right|^2-\frac{(c-k^2d)(\bar c-k^2 \bar d)}{1-k^2} + c\bar c -k^2d\bar d^2 \\ &= (1-k^2)\,\left|z -\frac{c-k^2d}{1-k^2}\right|^2 \\ &\quad\quad - \frac{\cancel{c \bar c}-k^2c\bar d-k^2 \bar c d+\bcancel{k^4 d \bar d}}{1-k^2}+\frac{\cancel{c\bar c} -k^2 c\bar c - k^2 d \bar d + \bcancel{k^4 d \bar d}}{1-k^2} \\[5px] &= (1-k^2)\,\left|z -\frac{c-k^2d}{1-k^2}\right|^2 - \frac{k^2}{1-k^2}\left(c\bar c +d\bar d-c\bar d-\bar c d\right) \\ &= (1-k^2)\,\left( \left|z -\frac{c-k^2d}{1-k^2}\right|^2 - \frac{k^2|c-d|^2}{(1-k^2)^2}\right) \end{align} $$

The geometric intuition of the latter equation becomes more obvious by defining the auxiliary variables $\,a=\cfrac{c+kd}{1+k}\,$ and $\,b=\cfrac{c-kd}{1-k}\,$. It can be easily verified that:

$$ a+b = 2\,\frac{c-k^2d}{1-k^2} \quad\quad\text{and}\quad\quad a-b=-2k\,\frac{c-d}{1-k^2} $$

Then the equation can be written as:

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

The above shows that the locus of $z$ is the circle having as diameter the two points $a,b$ which divide the segment $cd$ in ratio $k \ne 1$ internally, respectively externally