Apollonian Circles
Apollonian Circles are two families of circles such that every circles in the first family intersects orthogonally with every circles in the second family, and vice verse.
By Complex Number
Problem. Let $z, a, b \in \mathbb{C}$ correspond to points $P, A, B$ in the complex plane. Let $C_\lambda$ be the locus of $P$ defined by $|PA|/|PB| = \lambda$, where $\lambda$ is a fixed real positive constant. Show that $C_\lambda$ is a circle if $\lambda \not = 1$ and find its centre and radius.
Solution.
As $|z - a|^2 = (z - a)(\bar{z} - \bar{a})$ and $|z - b|^2 = (z - b)(\bar{z} - \bar{b})$, we have
\[(1 - \lambda^2)z\bar{z} - (a - \lambda^2b)\bar{z} - (\bar{a} - \lambda^2\bar{b})z + |a|^2 - \lambda^2|b|^2 = 0\]From that, we can see the centre $w_\lambda$ should be
\[w_\lambda = {a - \lambda^2b \over 1 - \lambda^2}\]By rearranging the terms, we have
\[\begin{align*} (z - w_\lambda)(\bar{z} - \bar{w_\lambda}) &= {1 \over (1 - \lambda^2)^2}\left( (1 - \lambda^2)\lambda^2|b|^2 - (1 - \lambda^2)|a|^2 + |a - \lambda^2b|^2\right) \\ &= {1 \over (1 - \lambda^2)^2} \left( \lambda^2|b|^2 + \lambda^2|a|^2 - \lambda^2(\bar{a}b + a\bar{b})\right) \\ &= \left( {\lambda \over 1 - \lambda^2} |a - b| \right)^2 \end{align*}\]Therefore,
\[r_\lambda = {\lambda \over 1 - \lambda^2} |a - b|\]When $\lambda = 1$, $|z - a| = |z - b|$, hence it becomes the perpendicular bisector of $a$ and $b$.
Problem. For the case $a = -b = p$, $p \in \mathbb{R}$, and for each fixed $\mu \in \mathbb{R}$, show that
\[S_\mu = \Set{ z \in \mathbb{C} : \left|z - i\mu\right| = \sqrt{p^2 + \mu^2}}\]is a circle passing through $A$ and $B$ with its centre on the perpendicular bisector of $AB$.
Solution.
By definition, $S_\mu$ is a circle with centre $w_\mu = i\mu$ and radius $r_\mu = \sqrt{p^2 + \mu^2}$. As $a = p$, $|a - i\mu| = |p - i\mu| = \sqrt{p^2 + \mu^2}$, the circle passes through $A$. Similarily, it passes through $B$.
As $|a - i\mu| = |b - i\mu|$, the centre of the circle $i\mu$ is always equidistant from $A$ and $B$, therefore it is the perpendicular bisector of $AB$.
Problem. Show that the circles $C_\lambda$ and $S_\mu$ intersect orthogonally for all $\lambda$ and $\mu$.
Solution.
As $a = -b = p$, we have
\[w_\lambda = {p - \lambda^2(-p) \over 1 - \lambda^2} = {1 + \lambda^2 \over 1 - \lambda^2}\,p\]and $|a - b| = 2p$, hence
\[r_\lambda = {2\lambda p \over 1 - \lambda^2}\]Suppose $C_\lambda$ and $S_\mu$ intersects at $z$, we have
\[|z - w_\lambda|^2 = r_\lambda^2 = \left({2\lambda p \over 1 - \lambda^2} \right)^2 \quad\text{and}\quad |z - w_\mu|^2 = r_\mu^2 = p^2 + \mu^2\]The square of the distance between the two centres is
\[\begin{align*} |w_\lambda - w_\mu|^2 &= \left| {1 + \lambda^2 \over 1 - \lambda^2}p - i\mu \right|^2 \\ &= \left({1 + \lambda^2 \over 1 - \lambda^2} \right)^2p^2 + \mu^2 \\ &= {4\lambda^2 + (1 - \lambda^2)^2 \over (1 -\lambda^2)^2}\,p^2 + \mu^2 \\ &= \left({2\lambda p \over 1 - \lambda^2} \right)^2 + p^2 + \mu^2 \\ &= r_\lambda^2 + r_\mu^2 \end{align*}\]Hence, for any $\lambda$ and $\mu$, by the converse of Pythagorean Theorem, the radii drawn to the points of intersection meet at right angle and the two circles $C_\lambda$ and $S_\mu$ intersects orthogonally.