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

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

Hi,

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!

$

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

$

I arrived at this point following this procedure:

$

(x-a)^2+(y-b)^2-r^2 = 0

$

Expanding the brackets ->

$
x^2+y^2-2ax-2by+a^2+b^2-r^2=0
$

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

$
x^2+y^2+2gx+2fy+g^2+f^2-r^2=0
$

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

$
g^2+f^2-r^2=c
$

$
x^2+y^2+2gx+2fy+c=0
$

gives the general circle equation (I believe!)

One representing each of the two intersecting circles ->

$
(1) x^2+y^2+2g_1x+2f_1y+c_1=0
$

$
(2) x^2+y^2+2g_2x+2f_2y+c_2=0
$

Subtract (2) from (1) ->

$
(2g_1-2g_2)x+(2f_1-2f_2)y+c_1-c_2=0
$

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

$
gx+fy+c=0
$

Subtract c and gx and then divide by f ->

$
y=(-gx-c)/f
$

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

$

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

$

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.

2. 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.