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 $AsB$}? 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.]
 A: The locus you are asking about is a subset of a cubic curve called an Apollonian strophoid (or strophoid of Apollonius).
The curve is the set of points $S$ for which either the internal or bisector of the angle $\angle ASB$ passes through $C$.  (The external bisector is the line through $S$ perpendicular to the internal bisector.)  As the strophoid $\Sigma$ comes in from infinity, $C$ is on the internal bisector but, as $\Sigma$ goes past $A$, $C$ is on the external bisector until $\Sigma$ passes $B$ after which $C$ is back on the internal bisector all the way to infinity.  So your locus is actually the strophoid minus the section between $A$ and $B$.
A GeoGebra applet that demonstrates the strophoid can be found here.  The strophoid is blue, the external bisector is green. You can move $S$ and $C$.  (Unfortunately the $S$ will jump around if you modify any of the points.  This is annoying but harmless.)  The strophoid is built using mathlove's equation.

There are two special cases where the cubic degenerates to a line $\ell$ and a circle $c$.   Both have been touched upon in other answers and comments for this question.

*

*when $C$ is on the perpendicular bisector of $A$ and $B$, line $\ell$ is the perpendicular bisector and $c$ is the circumcircle of $A,B,C$.  For points $S$ on $c$ it is the external bisector that goes through $C$.  (Because, if $M$ is the midpoint of $A,B$ then the triangle $\triangle ECS$ is a right angle.


*when $C$ is on the line $AB$, line $\ell$ is the line $AB$ and $c$ is an Apollonian circle of the segment $AB$.
For more about strophoids see the following references.
Gibert, Apollonian strophoids (part of a large catalog of so-called Triangle cubics)
mathcurve.com (Strophoid page)
Stachel, Strophoids, A Family Of Cubic Curves With Remarkable Properties (see Theorem 4).
A: Let $C(a,b),s(x,y)$ where $y\not=0$.
Also, let $U(u,0)$ be the intersection point of the angle bisector with $x$-axis.
Then, we have
$$\begin{align}&sA:sB=AU:BU
\\\\&\iff (1-u)\sqrt{(x+1)^2+y^2}=(1+u)\sqrt{(x-1)^2+y^2}
\\\\&\iff u=\frac{\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}}{\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}}\end{align}$$
So, $C$ is on the angle bisector ($sU$) if and only if
$$\begin{align}&b(x-u)=y(a-u)
\\\\&\iff (y-b)u=ay-bx
\\\\&\iff (y-b)\frac{\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}}{\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}}=ay-bx
\\\\&\iff (y-b)\bigg(\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}\bigg)
\\&\qquad\qquad=(ay-bx)\bigg(\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}\bigg)
\\\\&\iff (y-b-ay+bx)\sqrt{(x+1)^2+y^2}=(y-b+ay-bx)\sqrt{(x-1)^2+y^2}\end{align}$$
So, we have to have
$$(y-b-ay+bx)(y-b+ay-bx)\ge 0$$
under which we have
$$(y-b-ay+bx)^2\bigg((x+1)^2+y^2\bigg)=(y-b+ay-bx)^2\bigg((x-1)^2+y^2\bigg)$$
i.e.
$$\color{red}{bx^3-  a b x^2-  b x+(a^2 -b^2 +  1) x y -  a x^2 y +  b x y^2  -  a y^3+  a b y^2   -  a y     +  a b =0}$$

Examples :

*

*If $a=b=0$, then we get a line $x=0$.


*If $b=0$ with $a\not=0$, then we get a circle$$  \bigg(x-\frac{a^2+1}{2a}\bigg)^2 +  y^2   =\bigg(\frac{1-a^2}{2a}\bigg)^2$$under the condition that $-1\lt a\lt 1$ as jlammy commented.


*If $a=0$ with $b\not=0$, then we get a line $x=0$ or a circle $$x^2  +\bigg(y-\frac{b^2-1}{2b}\bigg)^2  =\bigg(\frac{b^2+1}{2b}\bigg)^2$$under the condition that $(y+bx-b)(y-bx-b)\ge 0$.
A: 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.
A: Let $M$ be the midpoint of $AB$.
$C$ lies on the perpendicular bisector of $AB$, hence $AC = BC$
$\angle AsC = \angle BsC$
$sC$ is common.
Hence either the angles $\angle sCA$ and $\angle sCB$ are equal or supplementary. If they're supplementary, then $ACB$ is a straight line, meaning $C$ is the midpoint of $AB$. We will consider this case at the end. Otherwise though, we have $\angle ACs = \angle BCs$, and so $\triangle ACs \equiv \triangle BCs$, so $As = Bs$, so $s$ must lie on the perpendicular bisector of $AB$, so $L$ = the perpendicular bisector of $AB$, as required.
Finally consider the case that $C$ is the midpoint of $BC$, and the angles are supplementary.
$$ \frac{\sin\angle sBC}{sC} = \frac{\sin\angle BsC}{BC} \text{(Sine law)}$$
$$ = \frac{\sin\angle AsC}{AC} \text{(AC = BC})$$
$$ = \frac{\sin\angle sAC}{sC} \text{(Sine Law)}$$
So we have $\sin\angle sBC = \sin\angle sAC$, so the two angles are either equal or supplementary. Since they are angles of a triangle, they cannot be supplementary, so they are equal, so $\triangle sAB$ is isoceles with base $AB$, so $As = sB$, so again we have that $s$ lies on the perpendicular bisector, as required.
So yes, the only points in $L$ are those on the perpendicular bisector of $AB$.
