来自付连续的问题
设m,n∈R,若直线(m+1)x+(n+1)y-2=0与圆(x-1)^2+(y-1)^2=1相切,则m+n的取值范围是?答案是m+n≤2-2倍根号2或m+n≥2+2倍根号2
设m,n∈R,若直线(m+1)x+(n+1)y-2=0与圆(x-1)^2+(y-1)^2=1相切,则m+n的取值范围是?
答案是m+n≤2-2倍根号2或m+n≥2+2倍根号2
1回答
2020-01-29 18:59