Question

直线(2k+1)x+(k+1)y=7k+4于圆(x-1)^2+(y+3)^2=25位置关系是

Answer 1

直线方程变形为 k(2x+y-7)+(x+y-4)=0,

令 2x+y-7=0,x+y-4=0,解得 x=3,y=1,

即 直线过定点P（3,1）,

将x=3,y=1代入圆的方程,则 (3-1)^2+(1+3)^2=20