曲线C2x^2+(y-r)^2=r^2 (r≠0)
C2为圆，圆心T(0,r),半径为|r|,
求PQ距离最小值，需先求|PT|的最小值
设P(x,y),则y=x²-1,x²=y+1 (y≥-1)
∴|PT|²=x²+(y-r)²
=y+1+y²-2ry+r²
=y²-(2r-1)y+r²+1
=[y-(r-1/2)]²+r²-(r-1/2)²+1
=[y-(r-1/2)]²+r+3/4
当r-1/2>-1即r>-1/2时，
y=r-1/2时，|PT|²min=r+3/4
由 r+3/4-r²>0
==>r²-r-3/4 -1/2