(1)直线L整理得到
x+y-4+m(2x+y-7)=0
解方程组
x+y-4=0
2x+y-7=0
得x=3 y=1
.'.直线L:(2m+1)x+(m+1)y-7m-4=0过定点(3,1)
(2)将(3,1)代入圆的方程,易知点在圆内。
设圆心到直线L的距离为d,弦长为2t,则满足
t²=r²-d²
要使t有最大值,则d取最小值,当直线过圆心时,d有最小值0,此时m=-1/3
要使t有最小值,则d取最大值
d²=a=(3m+1)^²/(5m²+6m+2)
=(9m²+6m+1)/(5m²+6m+2)
5am²+6am+2a=9m²+6m+1
(5a-9)m²+(6a-6)m+(2a-1)=0
这个方程有解必须
(6a-6)²-4(5a-9)(2a-1)≥0
a²-5a≤0
0≤a≤5
.'.dmax=√5
.'.2t=4√5
将a=5代入(5a-9)m²+(6a-6)m+(2a-1)=0,解得
m=-3/4
望采纳,谢谢
直线与圆交于两点,说明圆心到直线的距离小于半径,运用点到直线距离公式得: |2m+1+2(m+1)-7m-4|/√[(2m+1)^2+(m+1)^2]<5 即|-3m-1|/√[(2m+1)^2+(m+1)^2]<5 两边平方得 (3m+1)^2<25[(2m+1)^2+(m+1)^2 整理得 116m^2+144m+49>0 △=144^2-4*116*49<0 因此116m^2+144m+49>0恒成立因此不论m取什么实数,直线恒与圆相交于两点
很不错哦,你可以试下
f〃rㄎㄎhpsq∝长srㄎㄎy—∈t濞04717660262011-9-16 2:27:10
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9o後_忘却 正解!