Question

已知圆C：(x-1)^2+(y-2)^2=25,直线L：(2m+1)x+(m+1)y-7m-4=0

已知圆C：(x-1)^2+(y-2)^2=25,直线L：(2m+1)x+(m+1)y-7m-4=0

求证：1、直线l恒过定点。
2、判断直线l被圆C截得的弦何时最长、何时最短？并求截得的弦长最短时m的值以及最短长度？

最不明白的是第二问请详细些，谢谢！

Answer 1

（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，易知辩哗点在圆内。
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
这个方程有携运行解必须
(6a-6)²-4(5a-9)(2a-1)≥0
a²-5a≤0
0≤a≤5
.'.dmax=√5
.'.2t=4√5
将悄历a=5代入(5a-9)m²+(6a-6)m+(2a-1)=0，解得
m=-3/4
望采纳，谢谢

Answer 2

直线与圆交于两点，说明圆心到直肢昌线的距离小于半径，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恒成立樱唤因此不论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∝长脊饥凯srㄎㄎy—∈t濞04717660262011-9-16 2:27:10

Answer 3

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Answer 4

9o後_忘却 正解！