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 $x^2+y^2+2y-3=0$

Key fact: the perpendicular from the centre to a chord bisects that chord (by RHS congruence).

Rewriting the equation of the circle as $x^2+y^2-2y-3 = 0 \implies x^2+(y-1)^2 = 2^2$, the equation of the perpendicular is $y - 1 = -\frac{1}{m}x$. This intersects $y = mx - 3$ when $-\frac{1}{m}x +1 = mx - 3$, or:
$$-\left(\frac{1}{m}+m\right)x = -4 \implies x = \frac{4m}{1+m^2}$$
$$y = \frac{4m^2}{1+m^2} - 3 = \frac{4+4m^2-4}{1+m^2} - 3=1 - \frac{4}{1+m^2}.$$
Then substituting $m = \tan t$, we get:
$$x = \frac{4 \tan t}{\sec^2 t} = 4 \sin t \cos t = 2 \sin 2t$$
$$y = 1 - \frac{4}{\sec^2 t} = 1 - 4 \cdot \frac{1}{2}(1 + \cos 2t) = -2 \cos 2t-1$$
and hence $x^2+(y+1)^2 = 4$ or $x^2 + y^2 + 2y - 3 =0 $ in Cartesian form.

Note: The substitution $m = \tan t$ is useful as not only does it make use of the identity $1 + \tan^2 t = \sec^2 t$ to remove  the $1$, but $y = (\tan t)x$ is also the equation of the line which makes an angle of $t$ radians counter-clockwise from the $x$-axis.
