# Thread: Question regarding locus of midpoint of chord of circle.

1. ## Question regarding locus of midpoint of chord of circle.

Hi everyone. The full question is as follows:

A circle passes through $\displaystyle A(0,3), B(\sqrt{3},0), C(-\sqrt{3},0)$. 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 $\displaystyle y=mx-3$ of variable gradient $\displaystyle m$, cuts the circle ABC in two points $\displaystyle L$ and $\displaystyle M$ 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: $\displaystyle x^{2}+y^{2}-2y=3$

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

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

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

2. ## Re: Question regarding locus of midpoint of chord of circle.

Originally Posted by LimpSpider
I know that the answer to (d) is $\displaystyle (x^{2}+y^{2}+2y-3)(y+3)=0$.
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, $\displaystyle c=x_0^2+y_0^2$. Now, since (0, -4) ∈ line LM, $\displaystyle 0x_0 -4y_0=x_0^2+y_0^2$. This is the required equation in x0 and y0, only the locus has to lifted back up by 1.

3. ## Re: Question regarding locus of midpoint of chord of circle.

Thanks. It was quite clear. Although the y+3 is part of the answer given.