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Math Help - Question regarding locus of midpoint of chord of circle.

  1. #1
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    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)

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

    Quote Originally Posted by LimpSpider View Post
    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.
    Thanks from LimpSpider
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  3. #3
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    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.
    Last edited by LimpSpider; October 29th 2013 at 05:04 PM.
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