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.]

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 $(y+bx-b)(y-bx-b)\ge 0$.
