直线4x+3y+b=0与圆x2+y2-2x+2y+1=0有交点,求b的范围
问题描述:
直线4x+3y+b=0与圆x2+y2-2x+2y+1=0有交点,求b的范围
答
圆x2+y2-2x+2y+1=0可变形为(x-1)²+(y+1)²=1,圆心为(1,-1),
直线4x+3y+b=0即y= -4/3-b/3,斜率为-4/3,与圆有交点,不难得到切点为(1+0.8,-1+0.6),(1-0.8,-1-0.6),
则两切线方程为4*1.8+3*(-0.4)+b=0,4*0.2+3*(-1.6)+b=0
得4≥b≥-6