Trisectors in Apollonius Circle Find parametric equations of trisectors of angle $ APB  ( AP/PB= d_1/d_2= \lambda);\, AB, \lambda $ are constants of the Apollonius circle, and their envelopes.
Trisectors in Apollonius Circle
 A: Choose a coordinate system such that $A = (0,0)$ and $B = (1,0)$. Identify the Euclidean plane $\mathbb{E}^2$ with complex plane $\mathbb{C}$. We will use capital letters to denote points of $\mathbb{E}^2$ and corresponding lower case letters for the associated numbers in $\mathbb{C}$. 
We will only study the case $\mu \stackrel{def}{=} \lambda^{-1} = \frac{BP}{AP} = \left|\frac{1-p}{p}\right| \in (0,1)$. For such $\mu$, the locus of $P$ 
$$\mathcal{C} = \left\{\; P \in \mathbb{E}^2 : BP = \mu AP \;\right\}$$
will be a circle in $\mathbb{E}^2$ centered on $x$-axis and passing through the points $\left(\frac{1}{1\pm\mu},0\right)$.
Perform a inversion with respect to the unit circle in $\mathbb{E}^2$, i.e.
$$\mathbb{E}^2 \ni Z = (x,y)\;\mapsto\;\left(\frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}\right)
\in \mathbb{E}^2
\quad\leadsto\quad
\mathbb{C} \ni z = x+iy \;\mapsto\; \frac{1}{\bar{z}} \in \mathbb{C}$$
The image of $\mathcal{C}$ will be a circle centered on $(1,0)$ with radius $\mu$. This implies $\frac{1}{\bar{p}}$ and hence $p$ has a parametrization of the form
$$\frac{1}{\bar{p}} = 1 - \mu e^{-\theta i}\quad
\iff\quad p = \frac{1}{1 - \mu e^{\theta i}}
\quad\text{ for }\; \theta \in [-\pi,\pi]$$
For $P \in \mathcal{C}$ not lying on $x$-axis and point $Z \in \mathbb{E}^2 \setminus \{ P \}$, we have
$$\angle APZ = \arg \left(\frac{z-p}{0-p}\right) = \arg\left( 1 - \frac{z}{p} \right)$$
In particular for $Z = B$, we have
$$\angle APB = \arg\left(1 - \frac{1}{p}\right) = \arg\left(1 - (1-\mu e^{\theta i})\right) = \arg(\mu e^{\theta i}) = \theta$$
There are two trisectors $\ell_1$ and $\ell_2$ for $\angle APB$. 
For $k \in \{ 1, 2 \}$, set $\alpha = \frac{k}{3}$, $\beta = 1 -\alpha$ and $\gamma = \alpha-\beta$. Trisector $\ell_k$ satisfies the equation:
$$\begin{align} \arg\left( 1 - \frac{z}{p}\right) = \alpha\theta
\iff & \Im\left[ e^{-\alpha\theta i}\left(z(1-\mu e^{\theta i}) - 1 \right) \right] = 0\\
\iff & \Im\left[ z\left(e^{-\alpha\theta i} - \mu e^{\beta\theta i}\right)\right] + \sin(\alpha\theta) = 0\tag{*1a}
\end{align}
$$
To obtain an equation for the envelope of $\ell_k$, "differentiate" $(*1a)$
with respect to $\theta$ to get
$$
\Re\left[ z\left(\alpha e^{-\alpha\theta i} + \beta \mu e^{\beta\theta i}\right)\right] - \alpha\cos(\alpha\theta) = 0\tag{*1b}$$
Let $\eta = z e^{-\alpha\theta i}$, $(*1a)$ and $(*1b)$ can be rewritten as
$$\begin{array}{rcrcrcrc0}
( 1 - \mu e^{\theta i}) \eta 
&-& (1 - \mu e^{-\theta i}) \bar{\eta} 
&+& 2\sin(\alpha\theta)i &=& 0\\
(\alpha + \beta\mu e^{\theta i}) \eta 
&+& (\alpha + \beta \mu e^{-\theta i}) \bar{\eta} 
&-& 2\alpha\cos(\alpha\theta) &=& 0
\end{array}
$$
Eliminate $\bar{\eta}$ gives a parametrization of $\eta$ and hence $z$ as
$$
z = e^{\alpha\theta i}\left[
\frac{2\alpha\cos(\alpha\theta)(1 - \mu e^{-\theta i})
- 2\sin(\alpha\theta)i(\alpha + \beta\mu e^{-\theta i})}
{(1-\mu e^{\theta i})(\alpha + \beta\mu e^{-\theta i})
+(1-\mu e^{-\theta i})(\alpha + \beta\mu e^{\theta i})}
\right]
=  \frac{2\alpha - \mu e^{\gamma\theta i} - \gamma \mu e^{-\theta i}}{2(\alpha - \gamma \mu \cos\theta - \beta\mu^2)}
$$
Notice when $\alpha = \frac13$ or $\frac23$, $\gamma = -\frac13$ and $\frac13$.
We can parametrize $\alpha$ as $\frac{3+\epsilon}{6}$ and the trisectors $\ell_1$, $\ell_2$ corresponds to $\epsilon = -1$ or $+1$ respectively.
With this, we can simplify the equation of envlope of $\ell_k$ as
$$z = \frac{3 + \epsilon - 3\mu e^{\frac{\epsilon\theta}{3} i} - \epsilon\mu e^{-\theta i}}{3(1-\mu^2) + \epsilon (1 + \mu^2 - 2\mu\cos\theta)}$$
In terms of Cartesian coordinates, we have
$$
(x,y) = \left(
\frac{3( 1 - \mu \cos(\frac{\theta}{3})) + \epsilon(1-\mu \cos\theta)
}{3(1-\mu^2) + \epsilon (1 + \mu^2 - 2\mu\cos\theta)},
\frac{\epsilon\mu (\sin\theta - 3\sin\frac{\theta}{3})
}{3(1-\mu^2) + \epsilon (1 + \mu^2 - 2\mu\cos\theta)}
\right)\tag{*2}
$$
Following is an animation showing what happens when $\mu = 0.33$.
The light red and light blue lines are trisectors $\ell_1$ and $\ell_2$.
The red/blue curves are the corresponding loci.  
$\hspace0.75in$ 
For $\mu < \frac12$, both loci are connected curves with a cusp at $\theta = 0$. They behave qualitative like what presented above.
For $\mu > \frac12$, the loci for $\ell_2$ remains qualitatively the same.
However, the loci for $\ell_1$ behave differently, they split at $\theta = \cos^{-1}(2\mu - \frac{1}{\mu})$ into 3 connected components and each component has its own cusp. It is sort of hard to describe the configuration verbally. For those who are interested, it will be easier to plug formula $(*2)$ in your favorite CAS and look at the resulting graph.
