I'm looking for an analytic proof the statement for a Circle of Apollonius (I found a geometrical one already): If $\overline{AC}:\overline{BC}=s$, then $P \in k_s$. $s \in (0,1)$.
$k_s$ is the circle. I made the following drawing:
WLOG I can set $A=(0/0)$ and $B=(0/1)$. I really don't know how to go on. Perhaps someone can give a hint?
Greetings
Assume $s\in(0,1)$ and start with $$ \begin{align} s|A-C|&=|B-C|\\ s^2|A-C|^2&=|B-C|^2\\ s^2(|A|^2+|C|^2-2A\cdot C)&=|B|^2+|C|^2-2B\cdot C\\ 0&=(1-s^2)|C|^2-2(B-s^2A)\cdot C+|B|^2-s^2|A|^2\\ 0&=|C|^2-2\frac{B-s^2A}{1-s^2}\cdot C+\frac{|B|^2-s^2|A|^2}{1-s^2}\\ 0&=\left|C-\frac{B-s^2A}{1-s^2}\right|^2+\frac{|B|^2-s^2|A|^2}{1-s^2}-\left|\frac{B-s^2A}{1-s^2}\right|^2\\ \left|C-\frac{B-s^2A}{1-s^2}\right|^2&=\left|\frac{B-s^2A}{1-s^2}\right|^2-\frac{|B|^2-s^2|A|^2}{1-s^2}\\ \left|C-\frac{B-s^2A}{1-s^2}\right|^2&=\left(\frac{s}{1-s^2}|B-A|\right)^2\tag{1} \end{align} $$ Equation $(1)$ says that $C$ is on the circle with center $\dfrac{B-s^2A}{1-s^2}$ and radius $\frac{s}{1-s^2}|B-A|$.