# Given two points endpoints of circle chord find the locus of midpoint

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: $$x + my = 0$$ ?

• 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 $L$. – achille hui Aug 24 at 20:55
• @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 $x+my=0$. – Aleksandra Asanin Aug 24 at 21:19
• 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. $y = -\frac1m x \iff x + my = 0$. – achille hui Aug 24 at 21:26
• Well, I am not sure that the circle can rotate... But anyway thanks for your effort, maybe I figure out something – Aleksandra Asanin Aug 24 at 21:30
• 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. – achille hui Aug 24 at 21:33

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 $$-\dfrac 1m$$.

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

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.

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

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