What does $\arg\left(\frac{z-a}{z-b}\right)=\theta$ represent geometrically? ($a$ and $b \in \mathbb{C}$) Taking $z=x+iy$ and solving the equation algebraically shows that it is a circle but I cannot understand intuitively why this is so.
I would greatly appreciate a geometrical intuition behind this like there is with $|z-a|= r$ (postive numerical value), where $|z-a|$ represents the distance between $z$ and $a$, and $a$ being a fixed point, it represents a circle with radius $r$.
Also geometrical understanding of cases where $\arg\left(\frac{z-a}{z-b}\right)=\theta$ represent a pair of straight lines or other things. Thanks in advance!
Source: Problem-11 Tristan Needham Visual Complex Analysis page-46
 A: What is $\boldsymbol{\arg\left(\frac{z-a}{z-b}\right)}$?
$$
\begin{align}
\arg\left(\frac{z-a}{z-b}\right)
&=\arg\left(\frac{a-z}{b-z}\right)\\[3pt]
&=\arg(a-z)-\arg(b-z)
\end{align}
$$

That is, $\arg\left(\frac{z-a}{z-b}\right)$ is the counterclockwise angle from $b$ to $a$ as viewed from $z$.

For a given $\boldsymbol{\theta}$, what is the locus of $\boldsymbol{\arg\left(\frac{z-a}{z-b}\right)=\theta}$ ?
The Inscribed Angle Theorem says that the locus of points at which the angle from $b$ to $a$ is $\theta$ is an arc of a circle containing $a$ and $b$ so that the angle from $b$ to $a$ at the center of the circle is $2\theta$:

A: $$\arg\frac{z-a}{z-b}=\theta$$
Recall that $\arg w=\arctan\frac{\Im w}{\Re w}$
If $a=p+qi;\;b=r+si$ then 
$$\arg\frac{z-a}{z-b}=\arctan\frac{-p s+p y+q r-q x-r y+s x}{p r-p x+q s-q y-r x-s y+x^2+y^2}$$
Apply $\tan$ to both sides and get
$$\frac{-p s+p y+q r-q x-r y+s x}{p r-p x+q s-q y-r x-s y+x^2+y^2}=\tan\theta$$
Which can be rearranged and give the equation of a circle
$$x^2+y^2+hx+ky+j=0$$
where, called $\tan\theta=t$
$h= \left(-p+\frac{q}{t}-r-\frac{s}{t}\right)$
$k= \left(-\frac{p}{t}-q+\frac{r}{t}-s\right)$
$j=p r+\frac{p s}{t}-\frac{q r}{t}+q s$
