x=m±√[m(m+1)]因为：x2>0>x1
x1=m-√[m(m+1)]x=m+√[m(m+1)]
C(0,-m)
因为∠BAC = ∠BCO
tan∠BAC =OC/OA=m/(-x1)
tan∠BCO=x2/m
所以,m^2=-x1x2==-[m^2-m(m+1)]=m
m=0(不符合x2>0>x1,舍去),m=1
(1) y=x^2-2x-1
=(x-1)^2-2=(x-1+√2)(x-1-√2)
A(1-√2,0),B(1+√2,0),C(0,-1)
(2) 设圆D交x轴于另外一点O1
因为OA=√2-1,OO1=2√2,O1B=√2-1
OO1>O1B+OA,相当于：t=0时,FG>EF+GH
根据图像可知,随着t的增大,必存在这样的E(x3,t)
满足：FG=EF+GH
x3^2-2x3-1-t=0
设x32>0>x31,x32-x31=2FG,DG^2-(FG/4)^2=t^2
2-[(x32-x31)/4]^2=t^2
即:2-[4+4(1+t)16]=t^2
解方程得：t1=(√97-1)/8,t2