A Maths 問題

Let the equation of the circle be x^2+y^2+Dx+Ey+F=0
Put into the coordinates of the three points:
1+1+D+E+F=0
9+4+3D-2E+F=0
9+36-3D-6E+F=0

D+E+F= -2 _________(1)
3D-2E+F = -13________(2)
-3D-6E +F =-45 ______(3)

(2) -(1): 2D -3E = -11 ____(4)
(3)-(1): -4D-7E = -43 _____(5)

2*(4) + (5): -13E = -65 , E = 5
Put E=5 into (4): 2D -15 = -11, D=2

Put D=2 and E=5 into (1),
2+5+F = -2, F=-9

Hence the circle is x^2+y^2 +2x+5y -9 =0

https://lh3.googleusercontent.com/-0nIVPIb9gHE/Te7TFmJKHyI/AAAAAAAAF9s/9UN9iTvpqss/2510.JPG


圖片參考：https://lh3.googleusercontent.com/-0nIVPIb9gHE/Te7TFmJKHyI/AAAAAAAAF9s/9UN9iTvpqss/2510.JPG


2011-06-08 09:48:38 補充：
更正少加1 F=-9