Let $A$ and $B$ be two different points in the plane. Find the [Locus](https://en.wikipedia.org/wiki/Locus_(mathematics)) of all the points $C$ so $\angle ACB=\frac{2\pi}{3}$. 

What I tried to do:

[![enter image description here](https://i.stack.imgur.com/cxp0P.png)](https://i.stack.imgur.com/cxp0P.png)

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?