无论k取何值,直线(2k+1)x-(k+3)y-(k-2)=0总与圆x2+y2=r2(r>0)相交,则r的取值

无论k取何值,直线(2k+1)x-(k+3)y-(k-2)=0总与圆x2+y2=r2(r>0)相交,则r的取值范围是_______________.

答案

(,+∞)

解法一:直线方程变为(x-3y+2)+k(2x-y-1)=0,

令x-3y+2=0且2x-y-1=0,解得x=1,y=1.

故直线过定点P(1,1).

当P在圆内,则直线与圆总相交,

所以r2>12+12,即r>.

解法二:圆心(0,0)到直线的距离为

d==<r,

而(5d2-1)k2+(10d2+4)k+10d2-4=0,

Δ=(10d2+4)2-4(5d2-1)(10d2-4)≥0

*0≤d2≤2

*0≤d≤.

所以r>.

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