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Question: $A(2 \cos \theta_1 , 2 \sin \theta_1)$ 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: $x + my = 0$ ?

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  • $\begingroup$ Your question doesn't have enough info for an unique answer. If $L = |AB|$ is arbitrary, $M$ can be any point of the unit disk. If $L$ is fixed, the locus will be a circle whose radius depends of $L$. $\endgroup$ – achille hui Aug 24 at 20:55
  • $\begingroup$ @achillehui The only additional info I have is that the parametric equations of the circle are $x=2 \cos \theta$ and $y=2 \sin \theta$, as well as that we should suppose that the slope of AB is $\textbf{always}$ equal to $m$ (calculated above). Hope this helps. The answer in the book is $x+my=0$. $\endgroup$ – Aleksandra Asanin Aug 24 at 21:19
  • $\begingroup$ If you rotate everything by a suitable angle, you can make all chords horizontal (i.e. slope $=0$). The locus is the diameter parallel to $y$-axis. Rotate it back, the locus is a diameter with slope $-\frac1m$ (the locus is perpendicular to the chords). i.e. $y = -\frac1m x \iff x + my = 0$. $\endgroup$ – achille hui Aug 24 at 21:26
  • $\begingroup$ Well, I am not sure that the circle can rotate... But anyway thanks for your effort, maybe I figure out something $\endgroup$ – Aleksandra Asanin Aug 24 at 21:30
  • $\begingroup$ An alternative way to think about this is look at everything from a new coordinate system where the new $x$-axis is parallel to the chords. $\endgroup$ – achille hui Aug 24 at 21:33
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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 $-\dfrac 1m$.

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  • $\begingroup$ This looks like hell. Thanks a lot, I would never come to this point alone $\endgroup$ – Aleksandra Asanin Aug 25 at 17:26
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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.

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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: $\theta_{AB} = 90° + {\theta_1 +\theta_2 \over 2}$

$$m = \tan(\theta_{AB}) = \tan(90° + {\theta_1 +\theta_2 \over 2}) = -\cot({\theta_1 +\theta_2 \over 2}) $$

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