F.5 Maths 列式

Let equation of circle be
x^2 + y^2 + Dx + Ey + F = 0
For A(3,2)
9 + 4 + 3D + 2E + F = 0
3D + 2E + F = - 13..............(1)
For B(-5, 6)
25 + 36 - 5D + 6E + F = 0
-5D + 6E + F = - 61...........(2)
For C(-3,10)
9 + 100 - 3D + 10E + F = 0
-3D + 10E + F = -109...........(3)
(1) - (2) we get
8D - 4E = 48
2D - E = 12................(4)
(3) - (2) we get
2D + 4E = - 48
D + 2E = - 24............(5)
Sub (4) into (5) we get
D + 2(2D - 12) = - 24
5D - 24 = - 24
D = 0
so E = 2D - 12 = 0 - 12 = - 12
From (1), F = -13 - 3(0) - 2(- 12)
= -13 + 24 = 11
so the equation is
x^2 + y^2 - 12y + 11 = 0

設圓之方程為 x2 + y2 + Dx + Ey + F = 0 其中 D, E 和 F 為常數.
則代入 A, B 和 C 點之座標時:
A 點: 9 + 4 + 3D + 2E + F = 0
3D + 2E + F = - 13 ... (1)
B 點: 25 + 36 - 5D + 6E + F = 0
5D - 6E - F = 61 ... (2)
C 點: 9 + 100 - 3D + 10E + F = 0
3D - 10E - F = - 109 ... (3)
解 (1), (2) 和 (3) 可得出:
D = 109/5, E = 158/5 和 F = - 708/5
所以圓之方程為:
x2 + y2 + 109x/5 + 158y/5 - 708/5 = 0
5x2 + 5y2 + 109x + 158y - 708 = 0