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Thread: I'm struggling to rearrange to find x - simultaneous circle equations

  1. #1
    Apr 2009

    I'm struggling to rearrange to find x - simultaneous circle equations


    I need to rearrange the following in order to find x.

    I'm attempting to solve two circle equations simultaneously, I have subtracted one from the other and found y in terms of x, and substituted this back into one of the original equations in order to solve for x... but am not certain how to proceed!

    <br /> <br />
x^2+((-gx-c)/f)^2+g_1x+f_1((-gx-c)/f)+c_1=0<br /> <br />

    I arrived at this point following this procedure:

    <br /> <br />
(x-a)^2+(y-b)^2-r^2 = 0<br /> <br />

    Expanding the brackets ->

    <br />
x^2+y^2-2ax-2by+a^2+b^2-r^2=0<br />

    replacing (-a) with g and (-b) with f (apparently this is convention)->

    <br />
x^2+y^2+2gx+2fy+g^2+f^2-r^2=0<br />

    collect together the non-x and non-y terms ->

    <br />
g^2+f^2-r^2=c<br />

    <br />
x^2+y^2+2gx+2fy+c=0<br />

    gives the general circle equation (I believe!)

    One representing each of the two intersecting circles ->

    <br />
(1)     x^2+y^2+2g_1x+2f_1y+c_1=0<br />

    <br />
(2)     x^2+y^2+2g_2x+2f_2y+c_2=0<br />

    Subtract (2) from (1) ->

    <br />
(2g_1-2g_2)x+(2f_1-2f_2)y+c_1-c_2=0<br />

    Using g, f and c as the results of the subtractions ->

    <br />
gx+fy+c=0<br />

    Subtract c and gx and then divide by f ->

    <br />
y=(-gx-c)/f<br />

    Substituting this back into equation (1) gives the equation which I need to solve. (Also at top of post).

    <br /> <br />
x^2+((-gx-c)/f)^2+g_1x+f_1((-gx-c)/f)+c_1=0<br /> <br />

    I'm uncertain if I'm going about this in the best way, so if anybody has an easier alternative I would be very grateful.
    I cannot guarantee the circles will be orthogonal so I cannot use kite geometry.

    Any help/advice would be greatly appreciated.
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    What you have done looks correct to me. The final equation is a quadratic in x, which you can solve to find the x-coordinates of the two points of intersection. I doubt whether there is a neater way of doing this.
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