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Math Help - how to prove this?

  1. #1
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    how to prove this?

    Let two intersecting circles be represented by equations x^2 + y^2 + Ax + By + C =0 and x^2 + y^2 + ax + by + c = 0.

    For any number k that is not equal to -1, show that the equation of the circle through the intersection points of the two circles is x^2 + y^2 + Ax + By + C + k(x^2 + y^2 + ax + by + c) = 0.

    Conversely, also show that every such circle may be represented by such an equation with a suitable k.
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  2. #2
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    Re: how to prove this?

    First; every circle can be written in the form $x^2+y^2+ux+vy+w=0$ and every such equation represents a circle. Thus $x^2 + y^2 + Ax + By + C + k(x^2 + y^2 + ax + by + c) = 0$ is a circle iff $k\ne -1$.

    Also for the intersection point $(x,y)$ we have $x^2 + y^2 + Ax + By + C=0$ and $x^2 + y^2 + ax + by + c=0$ hence for such a point $x^2 + y^2 + Ax + By + C + k(x^2 + y^2 + ax + by + c) = 0$ so the intersection points of the two circles lie on the third circle.
    Last edited by zzephod; August 24th 2014 at 12:26 AM.
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  3. #3
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    Re: how to prove this?

    Your statement of the problem is not quite correct so I restated it. The previous response is not quite correct; here is a complete solution.

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