<s>[INST] Find locus of mid-point of chord of circle I am working through a pure maths book as a hobby and am struggling with the very last part of the question below. I give the whole of the question just in case it proves relevant to the answer.
A circle passes through the points A, B and C which have coordinates $(0,3),(\sqrt3,0)$ and $(-\sqrt3,0$ respectively.
Find (i) the equation of the circle , (ii) the length of the minor arc BC, (iii) the equation of the circle on AB as diameter.
And then follows this part which I cannot do:
A line $y=mx-3$ of variable gradient  m cuts the circle ABC in two points L and M. Find in cartesian form the equation of the locus of the mid-point of LM.
I have worked out the first parts to get:
(i) $x^2+y^2-2y-3=0$
(ii)$\frac{4\pi}{3}$
(iii)$x^2+y^2-3\sqrt x-3y=0; x^2+y^2+2y-3=0$
But I cannot get the very last part of (iii). I thought if I could find the coordinates of intersection I could then take the coordinates of the mid-point.
I have tried as follows:
Circle equation $x^2+y^2-2y-3=0 \rightarrow  y=\frac{x^2+y^2-3}{2}$
At intersection $\frac{x^2+y^2-3}{2}=mx-3$
I then try to solve for this but in no way can I get the answer which the book says is $x^2+y^2+2y-3=0$
 [/INST] We solve the system
$$\begin{cases}
x^2+y^2-2y-3=x^2+(y-1)^2-4=0\\
y=mx-3
\end{cases}$$
finding
$$x^2+(mx-3-1)^2-4=0$$
that is
$$(1+m^2)x^2-8mx+12=0\qquad(1).$$
Hence the locus of midpoints is given by
$$\frac{x_1+x_2}{2}=\frac{4m}{1+m^2},\quad\frac{y_1+y_2}{2}=\frac{4m^2}{1+m^2}-3=\frac{m^2-3}{1+m^2}.$$
The discriminant $\Delta$ of the quadratic equation (1) should be non-negative which yields the condition $|m|\geq \sqrt{3}$.
Note that for $m=0,\sqrt{3},-\sqrt{3}$, we get the points $(0,-3), (\pm\sqrt{3},0)$. Making a drawing, we guess that the locus is along a circle. Then we find the circle through the $3$ points (which are symmetric to $A,B,C$ with respect to the line $x=0$):
$$x^2+y^2+2y-3=0.$$
Finally, we verify that the whole locus of midpoints is along above circle:
$$x^2+y^2+2y-3=x^2+(y+1)^2-4=\left(\frac{4m}{1+m^2}\right)^2
+\left(\frac{2m^2-2}{1+m^2}\right)^2-4=0.$$
Notice that, because of the condition $|m|\geq \sqrt{3}$, the locus is not the whole circle, but just the $\color{blue}{\rm upper\ arc}$ with endpoints $(\pm\sqrt{3},0)$.

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