实数x y满足x^2+(y-1)^2=1,当x+y+d大于等于0恒成立时,d的取值范围

题目

实数x y满足x^2+(y-1)^2=1,当x+y+d大于等于0恒成立时,d的取值范围
答案是根号2-1到正无穷大的前闭后开区间,

答案

令x=cosa
则(y-1)²=1-cos²a=sin²a
所以y-1=sina
y=1+sina
所以x+y=sina+cosa+1=√2sin(a+π/4)+1
所以-√2+1<=√2sin(a+π/4)+1<=√2+1
即-√2+1<=x+y<√2+1
-√2-1<=-(x+y)<√2-1
-(x+y)最大值=√2-1
x+y+d>=0
d>=-(x+y)
所以只要d大于等于-(x+y)最大值即可
所以d>=√2-1