The locus of points $z$ which satisfy $|z - k^2c| = k|z - c|$, for $k \neq 1$, is a circle Use algebra to prove that the locus of points z which satisfy $|z - k^2c| = k|z - c|$, for $k \neq 1$ and $c = a + bi$ any fixed complex number, is a circle centre $O$.
Give the radius of the circle in terms of $k$ and $|c|$.
I squared both sides and got this:
 $$(k^2−1)x^2+(k^2−1)y^2+(a^2+b^2-k^2a^2-k^2b^2)k^2=0$$
I might have gone wrong somewhere though.
Edit. Never mind, I didn't go wrong.
$$(k^2-1)x^2+(k^2-1)y^2-(k^2-1)k^2a^2-(k^2-1)k^2b^2=0$$
$$x^2+y^2=k^2(a^2+b^2)$$
$$r^2=k^2(a^2+b^2)$$
$$r=k|c|$$

instead of squaring let me see if conjugates help.
$\begin{eqnarray}
0 = |z - k^2c|^2 -k^2 |z -c|^2 &=& (z - k^2 c)(\bar z - k^2 \bar c) -k^2(z-c)(\bar z - \bar c) \\
&=& z \bar z - k^2 c \bar z - k^2\bar c z + k^4 c \bar c -k^2(z \bar z - c \bar z -z  \bar c + c \bar c)  )\\
&=&(1-k^2)z \bar z - k^2(1-k^2)c\bar c \\
&=&(1-k^2)(|z|^2 - k^2|c|^2)
\end{eqnarray}$
$ \mbox{ if } k^2 \neq 1, \mbox{ then } |z|^2 = k^2|c|^2$
