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Math Help - intersecting circles

  1. #1
    Newbie FRANTICSTAR's Avatar
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    intersecting circles

    Hi I'd like to know how I find out where 2 circles would intersect

    I can show the following example:



    So far I have:

    Left circle:
    x = 250 + 200 * cos(a1)
    y = 350 + 200 * sin(a1)

    where a1 has 2 values where the angles where the circles intersect

    and the right circle:
    x = 400 + 200 * cos(a2)
    y = 250 + 200 * sin(a2)

    where a2 has 2 values where the angles where the circles intersect

    from that I have:

    250 + 200 * cos(a1) = 400 + 200 * cos(a2)
    350 + 200 * sin(a1) = 250 + 200 * sin(a2)

    Which is where I get stuck.
    Last edited by FRANTICSTAR; March 13th 2009 at 11:27 AM.
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  2. #2
    MHF Contributor
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    Hi

    Maybe it is more simple with Cartesian equations

    The first circle equation is (x-250)^2+(y-350)^2=200^2
    The second circle equation is (x-400)^2+(y-250)^2=200^2

    A point on the two circles satisfies both equations

    Subtracting gives
    300x-200y+350^2-400^2=0

    y=\frac32\:x-\frac{375}{2}

    Substituting in equation 2 gives a quadratic equation whose solutions are approximately
    x_1 = 225.97 => y_1 = 151.45
    x_2 = 424.03 => y_2 = 448.55
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  3. #3
    Newbie FRANTICSTAR's Avatar
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    Hi running-gag,

    This looks great, I've never heard of Cartesian equations, i'm a little confused how you get to:

    300x-200y+350^2-400^2=0

    and

    y = 3/2x - 375/2
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  4. #4
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by FRANTICSTAR View Post
    This looks great, I've never heard of Cartesian equations
    You have learned about parametric equations, but not Cartesian (rectangular) ones? I find that a bit hard to believe.

    300x-200y+350^2-400^2=0
    As running-gag said, you subtract:

    \begin{tabular}{cc}<br />
&$(x-250)^2+(y-350)^2=200^2$\\<br />
$-$&$(x-400)^2+(y-250)^2=200^2$\\\hline<br />
\end{tabular}

    which gives

    (x-250)^2+(y-350)^2-\left[(x-400)^2+(y-250)^2\right]=200^2-200^2

    \Rightarrow(x-250)^2+(y-350)^2-(x-400)^2-(y-250)^2=0

    \Rightarrow(x^2-500x+250^2)+(y^2-700y+350^2)-(x^2-800x+400^2)-(y^2-500y+250^2)=0

    \Rightarrow300x-200y+350^2-400^2=0.

    y = 3/2x - 375/2
    Continuing from the above,

    300x-200y+350^2-400^2=0

    \Rightarrow3x-2y+35^2-40^2=0

    \Rightarrow y=\frac{3x+35^2-40^2}2

    \Rightarrow y=\frac32x+\frac{35^2-40^2}2

    \Rightarrow y=\frac32x+\frac{25(7^2-8^2)}2

    \Rightarrow y=\frac32x+\frac{25(-15)}2

    \Rightarrow y=\frac32x-\frac{375}2.
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  5. #5
    Newbie FRANTICSTAR's Avatar
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    Cool I think that's starting to make some sense!

    I didn't know they were called parametric equations, I learnt those equations from programming my amastrad when I was 11.
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