The slope of the line of midpoints must be $-\dfrac 1m$ and the line must contain the origin. So the equation is $y = -\dfrac 1mx$ or $x+my=0$
You know that
\begin{align}
M_{1,2}
&= \left(\cos\theta_1 + \cos\theta_2 , \
\sin\theta_1 + \sin\theta_2 \right) \\
&= \left(2\cos\dfrac{\theta_1+\theta_2}{2}
\cos\dfrac{\theta_1-\theta_2}{2} , \
2\sin\dfrac{\theta_1+\theta_2}{2}
\cos\dfrac{\theta_1-\theta_2}{2} \right)
\end{align}
We also know that $M_0 =(0,0)$ is the midpoint of the line with slope $m$ that passes through the origin. The slope of the line through $M_0$ and $M_{1,2}$ is
$m' = \dfrac
{2\sin\dfrac{\theta_1+\theta_2}{2}\cos\dfrac{\theta_1-\theta_2}{2}}
{2\cos\dfrac{\theta_1+\theta_2}{2}\cos\dfrac{\theta_1-\theta_2}{2}}
= \tan \dfrac{\theta_1+\theta_2}{2}
$
You also know that
\begin{align}
m
&= \dfrac{\sin \theta_2 - \sin \theta_1}
{ \cos \theta_2 - \cos \theta_1} \\
&= \dfrac{ 2 \sin \dfrac{\theta_2 - \theta_1}{2}
\cos \dfrac{\theta_2 + \theta_1}{2}}
{-2 \sin \dfrac{\theta_2 - \theta_1}{2}
\sin \dfrac{\theta_2 + \theta_1}{2}} \\
&= -\cot \dfrac{\theta_2 + \theta_1}{2}
\end{align}
So, for all $\theta_1$ and $\theta_2$ that represent the angles corresponding to the endpoints of a line with slope $m$ intersecting the circle $x^2+y^2 = 4$, $m \cdot m' = -1$.
In other words, the locus of the bisectors is a line with slope $-\dfrac 1m$.