Expert: Sherman D. - 11/23/2007

Question
suppose we have two circle as
a1x^2 + b1y^2 + 2g1x + 2h1y + c1 = 0
a2x^2 + b2y^2 + 2g2x + 2h2y + c2 = 0
kindly please tell me how to proceed to the solution  

Answer
a1x^2 + b1y^2 + 2g1x + 2h1y + c1 = 0

(a1x^2 + 2g1x) + (b1y^2 + 2h1y) = -c1

a1(x^2 + (2g1/a1)x) + b1(y^2 + (2h1/b1)y) = -c1

a1(x^2 + (2g1/a1)x + (g1/a1)^2 - (g1/a1)^2) + b1(y^2 + (2h1/b1)y + (h1/b1)^2 - (h1/b1)^2) = -c1

a1((x + (g1/a1))^2 - (g1/a1)^2) + b1((y + (h1/b1)^2) - (h1/b1)^2) = -c1

a1(x + (g1/a1))^2 - (g1^2/a1) + b1(y + (h1/b1))^2 - (h1^2/b1) = -c1

a1(x + (g1/a1))^2 + b1(y + (h1/b1))^2 = (g1^2/a1) + (h1^2/b1) - c1

same method goes for the second circle, just replace the 1s with 2s.

it would be even easier if their were actual values.