Curve for constant angle for two fixed points Consider two points $A$ and $C$ separated by a distance $d$.  One can show through simple geometry that from any point $B$ on the semicircle having radius $d/2$ with $A$ and $C$ as specifying its diameter the angle $\angle ABC = 90^\circ$.  
What is the name (and functional form) of the equivalent curve when the angle is some arbitrary fixed angle $0 \leq \angle ABC \leq 180^\circ$?

 A: Hint. Build an isosceles triangle $ABC$ with basis $AC$ so that $\hat{B}$ has a pre-defined angle and consider the circle with center on $B$ and radius $BA$.
Full answer. Let $\alpha \in (0^\circ, 90^\circ)$ be a fixed angle.
There are precisely $2$ isosceles triangles $AB_iC$ with basis
$AC$ such that  $A\widehat{B}_iC = 2 \alpha$. Since $2\alpha < 180^\circ$,
it is possible to build  these triangles. Also, notice that one is the
reflection of the other by line  $AC$. Let us denote by $\Gamma_i$ the
circle centered on $B_i$ that has radius  $AB_i$. Notice that the
chord $AC$ breaks each circle $\Gamma_i$ into two arcs.  Using the
inscribed angle theorem,
we know that in one
of the arcs  of $\Gamma_i$ the angle that any point sees arc $AC$ is
precisely $\alpha$. On the  complementary arc of $\Gamma_i$, the angle
that any point sees $AC$ is $180^\circ 
- \alpha$, since for any inscribed quadrilateral the sum of opposing angles
is  $180^\circ$.

A: WLOG, assume that $A$ is at $(0, 0)$ and $C$ is at $(d, 0)$. We then need to find all points $B = (x, y)$ such that $\angle ABC = \theta$.
We have that $$\tan(\theta) = \frac{-yd}{x(x-d)+y^2}$$
For now, let's assume that $y > 0$. The other half of the answer will just be the reflection over the $x$-axis. We then have that $$x^2\tan(\theta)-dx\tan(\theta)+y^2\tan(\theta)+yd=0$$
As you can see, each half is part of a circle with radius $d\csc(\theta)/2$ centered at $(d/2, \pm d\cot(\theta)/2)$ going to the $x$-axis.
