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: $x + my = 0$ ?
 A: The parameters $\theta_1$ and $\theta_2$ serve no useful purpose other than allowing you to find a Cartesian equation for the circle, so now that you’ve done that, let’s forget about them. The endpoints of the chords are the intersections of the line $y=mx+b$ with the circle $x^2+y^2=4$, with fixed slope $m$ and variable $y$-intercept $b$. Some straightforward algebra yields the points $$\left({-mb\pm\sqrt{4(1+m^2)-b^2} \over 1+m^2}, {b\pm m\sqrt{4(1+m^2)-b^2} \over 1+m^2}\right)$$ and so $$M(b) = \left(-{mb\over 1+m^2}, {b\over 1+m^2}\right).$$ Eliminate $b$ to get a Cartesian equation for this curve. Note, though, that not all values of $b$ are valid—most lines won’t intersect the circle. For the line to have two intersection points with real coordinates, we must have $-2\sqrt{1+m^2}\lt b\lt 2\sqrt{1+m^2}$.  
Incidentally, this is a special case of a general theorem about midpoints of parallel chords of an ellipse.
A: 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 $-\dfrac 1m$.
A: Think of A, B on the circle (center C), at angle $\theta_1, \theta_2$
Thus, AB's midpoint M, inside circle, angle = ${\theta_1 +\theta_2 \over 2}$
AB perpendicular to MC, with angle to the x-axis: $\theta_{AB} = 90° + {\theta_1 +\theta_2 \over 2}$
$$m = \tan(\theta_{AB}) =  \tan(90° + {\theta_1 +\theta_2 \over 2})
= -\cot({\theta_1 +\theta_2 \over 2}) $$
