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.
$$(k^2-1)x^2+(k^2-1)y^2-(k^2-1)k^2a^2-(k^2-1)k^2b^2=0$$
$$x^2+y^2=k^2(a^2+b^2)$$
$$r^2=k^2(a^2+b^2)$$
$$r=k|c|$$

Hint: If you square both sides, and expand out (initially writing things in terms of your variables and their complex conjugates), you will get a lot of useful cancellation.
Here is the beginning of such a calculation, to help you out.
Squaring the left hand side yields:
$$(z-k^2c)(\overline{z}-k^2\overline{c})=|z|^2+k^4|c|^2-k^2(z\overline{c}+\overline{z}c).$$
Similarly, squaring the right hand side yields:
$$k^2(|z|^2+|c|^2-(z\overline{c}+\overline{z}c)).$$
Setting the two sides equal and rearranging, we have
$$(k^2-1)|z|^2=(k^4-k^2)|c|^2.$$
