# Question regarding locus of midpoint of chord of circle.

• Oct 28th 2013, 09:35 PM
LimpSpider
Question regarding locus of midpoint of chord of circle.
Hi everyone. The full question is as follows:

A circle passes through $A(0,3), B(\sqrt{3},0), C(-\sqrt{3},0)$. Find:
(a)The equation of the circle
(b) the length of the minor arc BC
(c) the equation of the circle on AB as diameter.
(d) 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 midpoint of LM.

I have answered A through C, but (d) eludes me, and I can't even begin.
The answer for (a) is: $x^{2}+y^{2}-2y=3$

For (b): $\frac{4\pi}{3}$

For (c): $x^{2}+y^{2}-\sqrt{3}x-3y=0$

I know that the answer to (d) is $(x^{2}+y^{2}+2y-3)(y+3)=0$. I also know that the line $y=mx-3$ "rotates" around the point $(0,-3)$

• Oct 29th 2013, 01:20 PM
emakarov
Re: Question regarding locus of midpoint of chord of circle.
Quote:

Originally Posted by LimpSpider
I know that the answer to (d) is $(x^{2}+y^{2}+2y-3)(y+3)=0$.

I am not sure you have have y + 3 as a factor. All points (x, -3) satisfy this equation, and they are not even inside the circle.

Here is one solution to (d). Let's lower the circle by 1 so that the center is at (0, 0). Suppose the middle of a chord LM has coordinates (x0, y0). Then the line from (0, 0) to (x0, y0) is perpendicular to LM, so the line LM has the equation xx0 + yy0 = c for some constant c. (If this is unclear, please say so.) Since (x0, y0) ∈ LM, $c=x_0^2+y_0^2$. Now, since (0, -4) ∈ line LM, $0x_0 -4y_0=x_0^2+y_0^2$. This is the required equation in x0 and y0, only the locus has to lifted back up by 1.