Circle

Let the equation of the circle be x2+ y2 + Dx + Ey + F = 0
Put x=0, y2 + Ey + F = 0 ... (*)
Let α and β be the roots of (*). Since AB = 2√5,
α - β = 2√5
(α - β)2 = 20
(α + β)2 - 4αβ = 20
E2 - 4F = 20 ... (1)

Put y=0, x2 + Dx + F = 0 ... (**)
Since the circle touches the x-axis, Δ of (**) = 0
D2 - 4F = 0 ... (2)
As well, the centre lies on the line 2x - y + 1 = 0. Hence,
2(-D/2) - (-E/2) + 1 = 0
-D + E/2 + 1 = 0
E = 2D - 2... (3)
(1) - (2): D2 - E2 = -20
D2 - (2D - 2)2 = -20 [from (3)]
-3D2 + 8D - 4 = -20
3D2 - 8D - 16 = 0
(3D + 4)(D - 4) = 0
D = 4　or　D = -4/3
For D=4,
E = 2(4) - 2 = 6
(4)2 - 4F = 0
F = 4
For D=-4/3,
E = 2(-4/3) - 2 = -14/3
(-4/3)2 - 4F = 0
F = 4/9
Hence, the equation of the circle is:
x2 + y2 + 4x + 6y + 4 = 0　or　 x2 + y2 - 4x/3 - 14y/3 + 4/9 = 0




2007-08-18 18:40:55 補充：
Line 4:| α - β | = 2√5Sorry for missing the absolute sign.