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

Equation of circle is $x^2+y^2-2y-3=0$
Equation of line is $y = mx - 3$
Substituting $y$ from equation of line into equation of circle, we get
$x^2 + (mx-3)^2 - 2 (mx-3) - 3= 0$
Simplifying, $(1+m^2) x^2 - 8 mx + 12 = 0$
As the line intersects the circle at two points, the sum of roots of the above equation is the sum of x-coordinates of the intersection.
So, $x_1 + x_2 = \cfrac {8m}{1+m^2}$
If the midpoint is $(h, k)$,
$h = \cfrac{x_1 + x_2}{2} = \cfrac{4m}{1+m^2} \tag1$
Now we know that, $k = mh - 3$ as $(h, k)$ is on the line $y = mx - 3$.
So, $ \ k = \cfrac{4m^2}{1+m^2} - 3 = \cfrac{m^2-3}{1+m^2} \tag2$
From ($1$) and ($2$), $h^2 + k^2 = \cfrac{16m^2 + m^4 - 6m^2 + 9}{(1+m^2)^2}$
$ = \cfrac{(1+m^2)(m^2 + 9)}{(1+m^2)^2} = \cfrac{m^2 + 9}{1+m^2} = \cfrac{(3m^2 + 3) - (2m^2 - 6)}{1+m^2}$
So, $h^2 + k^2 = 3 - 2k \implies h^2 + k^2 + 2k - 3 = 0$

Alternatively from ($2$), $m = \pm \sqrt{\cfrac{k+3}{1-k}}$. Plug $m$ into ($1$), square both sides and resolve. You get the locus.
