Given two points endpoints of circle chord find the locus of midpoint 
$A(2 \cos \theta_1 , 2 \sin \theta_1)$ and $B(2 \cos \theta_2 , 2 \sin \theta_2)$ are two end points of a variable chord $AB$ of the circle, and $M$ is the midpoint of the chord. Suppose that the slope of the chord $AB$ is $\textbf{always}$ equal to $m$. Find the equation of the locus of $M$.

While trying to solve the problem I found that:

*

*Equation of the circle is $x^2 + y^2 = 4$


*Coordinates of chord midpoint $M$ are $\left( \cos \theta_1 +  \cos \theta_2 , \sin \theta_1 + \sin \theta_2 \right)$


*Slope of chord AB is $m = 
\dfrac{\sin \theta_2 - \sin \theta_1}{ \cos \theta_2 - \cos \theta_1}$
What should I do to get the solution: $x + my = 0$ ?

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 $-\dfrac 1m$.
