A.Math circle

The intersections of the two circles :
x^2+y^2+2x-4y=0 ...(1)
x^2+y^2-2y=0...(2)
(1)-(2) :
2x - 4y + 2y = 0
x = y , sub to (2) :
x^2 + x^2 - 2x = 0
x(x - 1) = 0
x = 0 or x = 1
So the intersections points is (0,0) and (1,1)
Let the equation of the require circle be x^2 + y^2 + Dx + Ey + F = 0
Sub x = 0 , y = 0 to it :
F = 0
sub x = 1 , y = 1 to it :
1 + 1 + D + E + 0 = 0
D+E = - 2
E = - 2 - D
Now , the equation of the circle becomes x^2 + y^2 + Dx + (-2-D)y = 0 which touches the line y = x + 3 , sub to it :
x^2 + (x+3)^2 + Dx - (2+D)(x+3) = 0
x^2 + x^2 + 6x + 9 + Dx - (2+D)x - 3(2+D) = 0
2x^2 + 4x + 3-D = 0 , since the circle touches the line , so x only have one solution , so
△ = 4^2 - 4(2)(3-D) = 0
16 - 24 + 8D = 0 D = 1
E = - 2 - D = - 2 - 1 = - 3
Hence the equation of the circle is x^2 + y^2 + x - 3y = 0


2009-09-17 21:56:08 補充：
Corrections :

2x^2 + 4x + 3 - 3D = 0 , since the circle touches the line , so x only have one solution , so

△ = 4^2 - 4(2)(3 - 3D) = 0

16 - 24 + 24D = 0

D = 1/3

E = - 2 - D = -2 - 1/3 = - 7/3

Hence the equation of the circle is x^2 + y^2 + x/3 - 7y/3 = 0

2009-09-17 22:12:34 補充：
(The last 6 lines)
Sorry for my mistakes !!