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Math Help - Equation of a Circle

  1. #1
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    Equation of a Circle

    I'm having some difficulty with this question. I can get to a certain point, and then the line equations cancel each other out. Can anyone help me out, please?

    Many thanks.

    Q.
    Find the equations of the 2 circles which pass through the points (1, 2) & (-1, 4) & have a radius of length \sqrt{10}.

    Attempt: Midpoint of (1, 2) & (-1, 4): (0, 3)
    Slope: \frac{4-2}{-1-1}=-1. Perpendicular slope: 1
    Equation of bisector of (1, 2) & (-1, 4): y-3=1(x-0)\rightarrow x-y+3=0
    Define the equation of the circle as x^2+y^2+2gx+2fx+c=0. Subbing in (1, 2) yields: 2g+4f+c=-5...1, subbing in (-1, 4) yields: -2g+8f+c=-17...2
    Solving 1 & 2 as simultaneous equations yields: g - f = 3
    From the bisector equation, because x -y + 3 = 0 contains centre c as (-g, -f), we can say -g + f = -3
    Applying simultaneous equation approach to g - f = 3 & -g + f = -3: g/ f = 0...

    Ans.: (From text book): (x-2)^2+(y-5)^2=10 & (x+2)^2+(y-1)^2=10
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  2. #2
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    Re: Equation of a Circle

    Hint: The distance between the centre and the points it passes through is equal to the radius.
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  3. #3
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    Re: Equation of a Circle

    Am I on the right track with (x-h)^2+(y-k)^2=10
    For (1, 2): (1-h)^2+(2-k)^2=10
    For (-1, 4): (-1-h)^2+(4-k)^2
    Thus (1-h)^2+(2-k)^2=(-1-h)^2+(4-k)^2
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  4. #4
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    Re: Equation of a Circle

    Yes you are on the right track
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  5. #5
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    Re: Equation of a Circle

    Ok, I have it now. Thank you very much.
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  6. #6
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    Re: Equation of a Circle

    Here is the solution

    Equation of a Circle-2circles.png
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