关于直线与圆的方程

A交B不等于空集即相切或相交
则圆心到直线距离小于等于半径
|t-1+2|/√2<=√2
|t+1|<=2
-2<=t+1<=2
-3<=t<=1

A交B不等于空集,说明方程
x-y+2=0
(x-t)^2+(y-1)^2=2
有解]
(x-t)^2+(x+2-1)^2-2=0
x^2-2tx+t^2+x^2+2x+1-2=0
2x^2+(2-2t)x+t^2-1=0
判别式
(2-2t)^2-8(t^2-1)>=0
t^2+2t-3<=0
(t-1)(t+3)<=0
-3<=t<=1

b:圆心为：(t,1)
(t-1+2)的绝对值除以根号2<=根号2
-3<=t<=1

解：A交B不等于空集,说明方程 x-y+2=0 和(x-t)^2+(y-1)^2=2 联立有解,由A得y=x+2,代入B得 (x-t)^2+(x+2-1)^2-2=0
所以2x^2+(2-2t)x+t^2-1=0
Δ=b^2-4ac=(2-2t)^2-8(t^2-1)>=0
得-3<=t<=1

x-y+2=0
(x-t)^2+(y-1)^2=2