Finding the locus of all the points $C$ so $\angle ACB=\frac{2\pi}{3}$ Let $A$ and $B$ be two different points in the plane. Find the Locus of all the points $C$ so $\angle ACB=\frac{2\pi}{3}$. 
What I tried to do:

Also the center of the circle is in $O(0,0)$. We can find the distance of $CO$:
$$
CO=\sqrt{x_c^2+y_c^2}
$$
Also we know that $CO=AO=BO$. If $\angle ACB=\frac{2\pi}{3}$ then due to being an inscribed angle we conclude that $\angle AOB=\frac{2\pi}{3}$.
Now I'm stuck. I understand that I need to get some equation that contains only $x_c$ and $y_c$ in order to get the locus. I though of using the Law of cosines:
$$
AB^2=AO^2+BO^2-2AO\cdot BO\cdot\cos120=2CO^2+CO^2=3CO^2
$$
But how can I represent $AB$ with only $x_c$ and $y_c$ so I could use the distance theorem and solve it?
 A: You are right that you essentially want to figure out where the origin is relative to $A$ and $B$  (and its reflection across $\overline{AB}$). Here is the construction you are looking for.

Construct $C$ such that $\triangle ABC$ is equilateral.  Bisect both $\angle BAC$ and $\angle ABC$, and let $D$ be their point of intersection.  Note that $m\angle BAD=m\angle ABD=\frac\pi6$, so $m\angle ADB=\frac{2\pi}3$.  Draw the minor arc of a circle with center $D$ connecting $A$ and $B$.  Extend $\overline{CD}$ to intersect this arc at $E$, and draw another minor arc with center $E$ connecting $A$ and $B$.  Those two arcs (excluding $A$ and $B$ themselves) are the locus of all points $X$ such that $m\angle AXB=\frac{2\pi}3$.
A: Hint
I believe we need to fix constant values to $A(m,n);B(r,s)$
so that we can use cosine rule
Otherwise
If the gradient of $CA,BC$ are $m,n$ respectively
$$\tan\dfrac{2\pi}3=\pm\dfrac{m-n}{1+mn}$$
A: Let the coordinates of the points $A(x_a, y_a)$ and $B(x_b, y_b)$. Then, the corresponding midpoint of $AB$ is $M(x_m,y_m)=M(\frac{x_a+x_b}2, \frac{y_a+y_b}2)$ and the directional vector perpendicular to the chord $AB$ is $(y_b-y_a,-x_b+x_a)$. The center of the locus circle lies on the line passing through the midpoint $M$ and perpendicular to $AB$, which can be parametrized as,
$$(x,y) = (x_m,y_m) +t(y_b-y_a,-x_b+x_a)$$
Knowing that the triangle $AOM$ is an 30-60-90 right triangle, we have
$$OM = \frac12 AB \>\cot 60=\frac1{2\sqrt3}AB$$
where 
$$OM^2= t^2[(x_b-x_a)^2+(y_b-y_a)^2]$$
$$AB^2= (x_b-x_a)^2+(y_b-y_a)^2$$
Substitute OM and AB into above equation to get the center parameter $t=\pm\frac1{2\sqrt3}$. Thus, the center of the circle is
$$(x_0,y_0) = (\frac{x_a+x_b}2\pm\frac{y_b-y_a}{2\sqrt3},\> \frac{y_a+y_b}2\mp\frac{x_b-x_a}{2\sqrt3})\tag 1$$
and its radius is $R = \frac12AB\csc60$, or
$$R^2=\frac13[(x_b-x_a)^2+(y_b-y_a)^2]\tag 2$$
As a result, the equation of the locus, expressed in terms of known coordinates $(x_a, y_a)$ and $(x_b, y_b)$, is
$$(x-x_0)^2+(y-y_0)^2 = R^2$$
where the center $(x_0,y_0) $ and the radius $R$ are given by (1) and (2), respectively. Note that there are two circles as shown by the two centers in (1).
