函数与几何的问题!帮忙! 过程!

(x-1)^2+(y-1)^2=4
y=0,x=1±√3
Ax=1-√3,Bx=1+√3,Ay=By=0
|AB|=2√3 cosAPB=(OA^2+OB^2-AB^2)/2OAOB=-4/8=-1/2,APB=120
y=ax^2+bx+c
Ax+Bx=-b/a=2 b/a=-2
AxBx=c/a=-2,c/a=-2
1=a+b+c
b=c=-2a
a=-1/3,b=2/3=c
y=-x^2/3+2x/3+2/3
y=(-1/3)(x-1)^2+1
D(x',y') PD中点Q,Qx=(1+x')/2,Qy=(1+y')/2
Q是OC中点,C(x'',y'')
Qx=x''/2,Qy=y''/2,1+x'=x'',1+y'=y''
y'=-(x'-1)^2/3+1
(1+x'-1)^2+(1+y'-1)^2=4
x'^2+[(x'-1)^2/3+1]^2=4
9x'^2+[(x'-1)^2+9]=36
9x'^2+(x'-1)^4+18(x'-1)^2+81=36
9x^2+(x'-1)^4+18(x'-1)^2+45=0
x'^2,(x'-1)^4,18(x'-1)^2>0方程无解,不存在这样的D点为什么(x-1)^2+(y-1)^2=4？圆心(a,b)半径r的圆的方程：(x-a)^2+(y-b)^2=r^2