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Math Help - Find the equation of the circle that passes through three points.

  1. #1
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    Find the equation of the circle that passes through three points.

    A ( 3, 1) B ( 2,4) C (-1, 2)

    some things that go with this equation

    -midpoint
    -distance
    -perpendicular bisector
    -M1-M2= -1
    -Simultaneous equations

    this is what i have so far i don't know if its correct
    (3-a)2 + (1-b)2 = r2 those 2's r squared im new to this and i just plugged in for the rest.

    Correct me if im wrong please and help me out if you can
    thank you.
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  2. #2
    Super Member bigwave's Avatar
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    Quote Originally Posted by Pre-Calc View Post
    A ( 3, 1) B ( 2,4) C (-1, 2)

    some things that go with this equation

    -midpoint
    -distance
    -perpendicular bisector
    -M1-M2= -1
    -Simultaneous equations

    this is what i have so far i don't know if its correct
    (3-a)2 + (1-b)2 = r2 those 2's r squared im new to this and i just plugged in for the rest.

    Correct me if im wrong please and help me out if you can
    thank you.
    you are doing fine...

    the standard equation of a circle is

    \left(x-h\right)^2 +\left(y-k\right)^2 = r^2

    from your equation a is h and b is K, it would probably be better to use the standard equation since that is what is commonly used.

    first find the center of the circle by doing a simultaneous equations of the 2 perpendicular bisectors
    of the lines AB and BC
    then you have the center C(h,k)

    to find the perpendicular bisector of line AB

    \sqrt{\left(x-3\right)^2+\left(y-1\right)^2}<br />
=\sqrt{\left(x-2\right)^2+\left(y-4\right)^2}<br />
    squaring both sides yields
    \left(x-3\right)^2+\left(y-1\right)^2=\left(x-2\right)^2+\left(y-4\right)^2<br />
    next expand
    x^2-6x+9+y^2-2y+1=x^2-4x+4+y^2-8y+16

    x-3y=-5

    continue with BC then find H,K by simultanuous equation
    Last edited by bigwave; January 6th 2010 at 12:24 AM. Reason: added info
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  3. #3
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    Thank you I understand it now.
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