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
