# Analytic proof for Circles of Apollonius

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

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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 $\frac{s}{1-s^2}|B-A|$.
@ulead86: I don't understand. Equation $(1)$ says that all $C$ whose distance from $B$ is $s$ times the distance from $A$ lie on the given circle. Did you have a different description of $k_s$? – robjohn Jun 12 '12 at 11:39
@ulead86: $\overline{AC}$ is a line segment. In context, sometimes people loosely equate $\overline{AC}$ and $\left|\overline{AC}\right|$. However, $A-C$ is the vector in the direction of $A$ from $C$ whose magnitude is $|A-C|=\left|\overline{AC}\right|$ – robjohn Jun 14 '12 at 15:04
@ulead86: Pretty straightforward computation: \begin{align} &\left|\frac{B-s^2A}{1-s^2}\right|^2-\frac{|B|^2-s^2|A|^2}{1-s^2}\\ &=\frac{\color{red}{|B|^2}-2s^2B\cdot A+\color{red}{s^4|A|^2}}{(1-s^2)^2}- \frac{\color{red}{|B|^2}-s^2|A|^2-s^2|B|^2 +\color{red}{s^4|A|^2}}{(1-s^2)^2}\\ &=\frac{s^2|B|^2-2s^2B\cdot A+s^2|A|^2}{(1-s^2)^2}\\ &=\frac{s^2}{(1-s^2)^2}|B-A|^2\\ &=\left(\frac{s}{1-s^2}|B-A|\right)^2 \end{align} – robjohn Jun 15 '12 at 13:39
@ulead86: points and vectors are both denoted as elements of $\mathbb{R}^n$: an array of $n$ elements of $\mathbb{R}$. As such, they can be added or subtracted; they can be multiplied or divided by real numbers; etc. The difference is how they behave under a change of coordinates (translation). If the points $A$ and $B$ are translated, $P=\frac{B-s^2A}{1-s^2}$ will be translated by the same amount. Therefore, $P$ represents a point. On the other hand $B-A$ does not change, so it represents a vector. Thus, a vector is the difference of two points, and a point plus a vector is another point. – robjohn Jun 18 '12 at 9:54
@ulead86: and yes, although $B$ is not a vector, we set $|B|=|B-0|$, where $0$ is the origin. – robjohn Jun 18 '12 at 9:57