x^2+y^2+x-6y+m=0与 x+2y-3=0交于p q,op垂直pq,求m
问题描述:
x^2+y^2+x-6y+m=0与 x+2y-3=0交于p q,op垂直pq,求m
答
将圆方程化简为标准式有: [x+(1/2)]^2+(y-3)^2=(37-4m)/4……………………………(1) 所以,圆心坐标为(-1/2,3) 联立直线与圆方程得到: x^2+x+y^2-6y+m=0 x+2y-3=0 ===> (2y-3)^2-(2y-3)+y^2-6y+m=0 ===> 4y^2-12y...