x^2+(y-2)^2=1
x^2=1-(y-2)^2＞0
(y-2)^2≤1
-1≤y-2≤1
1≤y≤3
P(x,y)到原点距离：
|OP| = √{x^2+y^2} = √{1-(y-2)^2+y^2} = √(4y-3)
1≤y≤3
4≤4y≤12
1≤4y-3≤9
1≤√(4y-3)≤3
1≤|OP|≤3