math-equation of circle

Write the standard form of the equation of the circle that passes through the points at (4,5) (-2, 3), and (-4,-3).
a. (x-5)^2+(y+4)^2=49
b. (x-3)^2+(y+2)^2=50
c. (x+4)^2+(y-2)^2=36
d. (x-2)^2+(y+2)^2=25

I'm having trouble with solving for the variables...after I make the system of equations, I use elimination and wind up with odd numbers (1/9 for E, 307/9 for F).
If somebody who is good at this could show me their steps it would really help me understand what I'm doing wrong.
Thanks so much in advance, I really appreciate the help! MUCH needed!

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  1. Since this is multiple choice, the quickest way to determine the correct solution would be to plug in the given coordinates into each of the solutions.

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  2. Well, call center at (xc,yc) and radius r
    then
    (x-xc)^2 + (y-yc)^2 = r^2
    distance from center to those points is the same = r
    (4-xc)^2 + (5-yc)^2 = r^2
    (-2-xc)^2 + (3-yc^2) = r^2
    (-4-xc)^2 + (-3-yc^2) = r^2

    16-8xc+xc^2+25-10yc+yc^2 = r^2
    4+4xc+xc^2+9-6yc+yc^2 = r^2
    16+8xc+xc^2+9+6yc+yc^2 = r^2

    -8xc+xc^2-10yc+yc^2 = r^2-41
    +4xc+xc^2-6yc+yc^2 = r^2-13
    +8xc+xc^2+6yc+yc^2 = r^2-25

    12 xc +4 yc = 28 changing sign of first and adding to second
    4xc +12 yc = -12 changing sign of second and adding to third

    12 xc +4 yc = 28
    12 xc +36 yc = -36
    -32yc = 64
    yc = -2

    ok, take it from there

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  3. of course it has to be b or d, but solve for xc to tell
    4 xc + 12(-2) = -12
    4 xc = 12
    xc = 3
    so center at
    (3,-2)
    and of form
    (x-3)^2 +(y+2)^2 = r^2

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  4. Thanks Damon!!
    This helped a lot...I was totally backwards on some things.

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