Find locus of mid-point of chord of circle

Question

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

Answer 1

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, $h^2 + k^2 = 3 - 2k \implies h^2 + k^2 + 2k - 3 = 0$

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

Answer 2

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 $(\pm\sqrt{3},0)$.

Answer 3

Let $O=(0,1)$ be the center of circle $ABC$, and $P=(0,-3)$ the common point of the pencil of lines with variable slope. If $N(x,y)$ is the midpoint of a chord $LM$, then $ON\perp LM$, i.e. $\angle ONP=90°$.

Hence the locus of $N$ is the locus of those points which see segment $OP$ under a right angle, that is a (part of a) circle of diameter $OP$.

Answer 4

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.

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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.