三題關於圓方程的問題..........

1
令圓方程x^2+y^2+Dx+Ey+F=0
代三點
F=0
9+3D+F=0
4-2E+F=0
得D=-3,E=2,F=0
圓的方程x^2+y^2-3x+2y=0
2
令圓方程x^2+y^2+Dx+Ey+F=0
代三點(-2,1),(-a,0),(0,a)
5-2D+E+F=0...(1)
a^2-aD+F=0...(2)
a^2+aE+F=0...(3)
aD=-aE
a(D+E)=0
D+E=0
代入(1) 5-2D+E+F=0=>F=2D-E-5=3D-5
此時圓的方程x^2+y^2+Dx-Dy+3D-5=0
代x=0﹐y^2-Dy+3D-5=0
判別式是0,D^2-12D+20=0
D=2或10
圓的方程x^2+y^2+2x-2y+1=0 或 x^2+y^2+10x-10y+25=0
3 設圓心M(h,k)
2h+k=0
k=-2h
因AM^2=BM^2
(h-2)^2+(k+1)^2=(h)^2+(k+3)^2
(h-2)^2+(-2h+1)^2=(h)^2+(-2h+3)^2
h^2-4h+4+4h^2-4h+1=h^2+4h^2-12h+9
5-8h=-12h+9
4h=4
h=1
k=-2
MB^2=(1)^2+(-2+3)^2=2
圓的方程 (x-1)^2+(y+2)^2=2
x^2+y^2-2x+4y+3=0


圓的方程x^2+y^2-3x+2y=0

D^2-12D+20=0
(D-10)(D-2)=0 (因式分解)

(D-10)=0 或 (D-2)=0 (一係(D-10)=0, 一係 (D-2)=0 )

D=10 或 D=2

我想問下"判別式是0,D^2-12D+20=0" 呢部點計出黎架????