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Let $c,d$ be different complex numbers and $k>0$. Show that if $k\neq 1$ then set $\{z \in \mathbb C: |z-c | = k|z-d|\}$ is a circle. Moreover, prove that if $z_0$ is centre of this circle then $z_0$ lies on the line between $c$ and $d$. Conclude, that $|z_0-c|\cdot|z_0-d| = r^2$.

Part $|z_0-c|\cdot|z_0-d| = r^2$ is trivial, since $z_0$ is centre then distances $|z-c|$ and $|z-d|$ are both equal to $r$.

For the first part, do I have to use the notation $z = x + y\cdot i$? Is it possible to deduce the "shape" of this set from some geometric interpretation?

To show that $z_0$ lies on the line between $c$ and $d$ I assume that we have to use form $z(t) = c + t\cdot(d-c)$, but still don't know how.

Any hints please?

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    Have a look at https://en.wikipedia.org/wiki/Circles_of_Apollonius. Possible duplicate of http://math.stackexchange.com/questions/582274/general-formula-for-loci-of-a-circle-in-the-complex-plane or http://math.stackexchange.com/questions/520469/apollonius-circle-its-radius-and-center. – 2017-02-28

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