Question

对任意实数k,圆C:x^2+y^2-6x-8y+12与直线l:kx-y+3=0的位置关系是

Answer 1

x^2+y^2-6x-8y+12＝0

→(x-3)^2+(y-4)^2＝(√13)^2.

圆心为(3,4),半径为R＝√13.

直线l与圆心(3,4)的距离

d＝|3k-1|/√(k^2+1)

→(d^2-9)k^2+6k+d^2-1＝0.

△＝36-4(d^2-9)(d^2-1)≥0,

解得,0