TITLE: Let $A=(-1,0),B=(1,0), C$ be points in $\mathbb{R}^2.$ What is the locus of points $\{s\in\mathbb{R}^2:C$ lies on the angle bisector of $AsB$}?
QUESTION [5 upvotes]: Let $A=(-1,0),\ B=(1,0),\ C$ be points in $\mathbb{R}^2.$ What (shape(s)) is the locus of points $L(C)=\{s\in\mathbb{R}^2:C$ lies on the angle bisector of $AsB$ } ?
Obviously the set $L$ depends on where $C$ is, which is why I've written $L$ as a function of $C.$
If $C$ is the midpoint of $AB$, or on the perpendicular bisector of $AB$, it is not clear to me that $L= \{$ the perpendicular bisector of $AB\},$ because maybe $L$ contains more points. [Edit: Karan Elangovan has answered this part of the question.]

REPLY [3 votes]: Geometric Construction for $\boldsymbol{C}$ off the Axes
Here is a geometric construction for the points of the locus.
We are given the points $A=(-1,0)$, $B=(1,0)$, and $C$. For any $\lambda\in\mathbb{R}$, define
$$
\begin{align}
P&=\frac{A+C}2+\lambda\rho(C-A)\tag{1a}\\
Q&=\frac{B+C}2+\lambda\rho(B-C)\tag{1b}
\end{align}
$$
where $\rho(v)=v\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ is $v$ rotated by $\frac\pi2$ counter-clockwise.
Then define two circles: one centered at $P$ passing through $A$ and $C$, and another centered at $Q$ passing through $B$ and $C$. Note that we are simply constructing two similar isosceles triangles: $\triangle APC$ and $\triangle BQC$.
By construction, both circles pass through $C$. Let the other point of  intersection of these circles be $D$.

By construction, $\angle APC=2\tan^{-1}\left(\frac1{2\lambda}\right)=\angle BQC$. Thus, by the Inscribed Angle Theorem,
$$
\angle ADC=\tan^{-1}\left(\frac1{2\lambda}\right)=\angle BDC\tag2
$$
The construction above works because $A$ is outside the circle containing $B$ and $B$ is outside the circle containing $A$. This construction also works when $A$ is inside the circle containing $B$ and $B$ is inside the circle containing $A$

Again, $\angle APC=2\tan^{-1}\left(\frac1{2\lambda}\right)=\angle BQC$. Thus, by the Inscribed Angle Theorem,
$$
\angle ADC=\tan^{-1}\left(\frac1{2\lambda}\right)=\angle BDC\tag3
$$
However, when only one of $A$ or $B$ is inside the circle containing the other, we have the following situation

Note that by the Inscribed Angle Theorem, $\angle BDC=\frac12\angle BQC$; however, $\angle ADC=\pi-\frac12\angle APC$.
$\angle ADC$ becomes supplementary to half its associated central angle, $\angle APC$, when $D$ passes through $A$; that is when $A$ changes whether it is inside or outside of the circle containing $B$. This happens when
$$
\lambda=\lambda_A\equiv\frac{(A-C)\cdot(B-A)}{2\rho(A-C)\cdot(B-C)}\tag4
$$
$\angle BDC$ becomes supplementary to half its associated central angle, $\angle BQC$, when $D$ passes through $B$; that is when $B$ changes whether it is inside or outside of the circle containing $A$. This happens when
$$
\lambda=\lambda_B\equiv\frac{(C-B)\cdot(B-A)}{2\rho(A-C)\cdot(B-C)}\tag5
$$
Which angle, $\angle ADC$ or $\angle BDC$, is supplementary to half its associated central angle switches when $D$ passes through $C$; that is, when the circles are internally tangent. This does not happen when the circles are externally tangent since then neither $A$ nor $B$ is inside the circle containing the other.
If we set
$$
\begin{align}
\mu&=\frac{(A-C)\cdot(B-C)}{\rho(A-C)\cdot(B-C)}\tag{6a}\\[3pt]
\sigma&=\operatorname{sign}(\rho(A-C)\cdot(B-C))\tag{6b}
\end{align}
$$
then the circles are internally tangent when
$$
\lambda=\frac{\mu-\sigma\sqrt{\mu^2+1}}2\tag7
$$
and externally tangent when
$$
\lambda=\frac{\mu+\sigma\sqrt{\mu^2+1}}2\tag8
$$
Here is an animation of $D$ (red point) traversing the locus of points (blue curve) where $C$ lies on the bisector of $\angle ADC$. The red curve is where one or the other inscribed angles is supplementary to half of its associated central angle.


Locus for $\boldsymbol{C}$ on the Axes
If $C$ is on the $y$-axis, then the locus is all of the $y$-axis except $C$.
If $C$ is on the $x$-axis, then the locus is all of the $x$-axis with the segment from $(-1,0)$ to $(1,0)$ and $C$ removed.
If $C=(0,0)$ then the locus is the union of the $x$-axis and $y$-axis with the segment of the $x$-axis from $(-1,0)$ to $(1,0)$ removed.