已知直线L:(m+1)x+(m-1)y-3m-1=0 圆C：x^2+y^2=9 (1)判定直线L与圆的位置关系 (2)若L与C交于AB两点 求当AB

已知C:x^2+y^2-2y-4=0,l:mx-y+1-m=0,(1)判断直线l和圆C位置关系~

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有条件，可得直线l过圆心
因此m=0.

(m+1)x+(m-1)y-3m-1=0
(m+1)x+(m-1)y=3m+1
(x+y)m+x-y=3m+1
x+y=3
x-y=1
则x=2，y=1
直线L:(m+1)x+(m-1)y-3m-1=0恒过(2,1)
圆C：x^2+y^2=9
则直线与圆C相交

设AB的中点为N
当圆C：x^2+y^2=9 的圆心O与N的连线ON⊥AB时，AB最小
则N点即(2,1)
向量ON;(2,1)
直线的方向向量就是(-1,2)
直线方程为2x+y-5=0

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分析：圆心（0,0）半径3.和直线L的方程可得d,d与r的关心进行分类讨论，L交AB，则L为圆C直径

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溪湖区13475318018： 已知圆C:(x - 1)2+(y - 2)2=25 及直线L:(2m+1)x+(m+1)y=7m+4 - ？
韩购帕朱： (1)直线的方程可以化简为:x+y-4+m(2x+y-7)=0 这表示过两直线x+y-4=0与2x+y-7=0交点的直线系;解方程组得交点为(3,1) 所以直线L:(2m+1)x+(m+1)y=7m+4恒过定点(3,1) (2)因为点(3,1)在已知圆的内部,所以当直线L过圆心时截得的弦为直径=10最大;当直线L与取最大值时的直线垂直时截得的弦最短;由于圆心到定点的距离为√5 所以最短弦长=2√(25-5)=4√5 此时最长弦的斜率=(2-1)/(1-3)=-1/2 所以最短弦的斜率=2;所以 - (2m+1)/(m+1)=2 ; m=-3/4

溪湖区13475318018： 已知圆C:(x - 1)2+(y - 2)2=25,直线l:(2m+1)x+(m+1)y - 7m - 4=0.(1)求证:直线l恒过定点;(2)求 - ？
韩购帕朱： (1)证明:直线l:(2m+1)x+(m+1)y-7m-4=0可化为m(2x+y-7)+(x+y-4)=0 令 2x+y-7=0 x+y-4=0 ,解得 x=3 y=1 ∴直线l恒过定点A(3,1) (2)解:直线l被圆C截得的弦长的最小时,弦心距最大,此时CA⊥l ∵圆C:(x-1)2+(y-2)2=25,圆心(1,2),半径为5 ∴CA的斜率为2-1 1-3 =-1 2 ,∴l的斜率为2 ∵直线l:(2m+1)x+(m+1)y-7m-4=0的斜率为-2m+1 m+1 ∴-2m+1 m+1 =2 ∴m=-3 4 ∵|CA|= 4+1 = 5 ∴直线l被圆C截得的弦长的最小值为2 25-5 =4 5 .

溪湖区13475318018： 已知圆C:(x - 2)²+(y - 2)²=9,直线l:(2m+1)x+(m+1)y - 7m - 4=0,m属 - ？
韩购帕朱： 直线l:(2m+1)x+(m+1)y-7m-4=0,m属于R,变为m(2x+y-7)+x+y-4=0,由2x+y-7=0,x+y-4=0解得x=3,y=1,∴l过定点A(3,1),AC=√5∴点A在圆C内部,∴不论m取什么实数,直线l与圆恒交于两点.②圆心C(2,2)到l的距离d=|2(2m+1)+2(m+1)-7m-4|/√[(2...

溪湖区13475318018： 已知m为实数,直线l:(2m+1)x+(1 - m)y - (4m+5)=0,P(7,0),则P到L的距离的取值范围为().写出过程 - ？
韩购帕朱： 以m为主元,直线l:(2m+1)x+(1-m)y-(4m+5)=0可以变形为(2x-y-4)m+(x+y-5)=0,不论m取何值,直线l都过两直线2x-y-4=0和x+y-5=0的交点Q(3,2),且|PQ|=2√5,因此,点P(7,0)到直线l的r的距离的取值范围是[0,2√5]

溪湖区13475318018： 已知直线(m+1)x+(1 - m)y - m - 3=0 求原点(0,0)到直线的距离的最大值 - ？
韩购帕朱： 解:把直线方程改写为:(x-y-1)m+(x+y-3)=0 令:x-y-1=0 x+y-3=0 解得:x=2, y=1.∴该直线恒过定点P(2, 1) 数形结合可知,原点到直线的最大距离,就是点P到原点的距离=√5

溪湖区13475318018： 已知直线L:(2m+1)x+(m - 1)y - 3m=0,圆C:x^2+y^2=4 求若圆上存在两点关于L对称,求m的值 - ？
韩购帕朱：[答案] 有条件,可得直线l过圆心 因此m=0.

溪湖区13475318018： 已知直线l:(2m+1)x+(m+1)y=7m+4与圆C:(x - 1)²+(y - 2)²=25.(1)求证:直线l与圆C总相交;(2)求出相交弦长的最小值及对应的m值. - ？
韩购帕朱：[答案] (2m+1)x+(m+1)y=7m+4 转化:(2x+y-7)m+(x+y-4)=0 联立:2x+y-7=0,x+y-4=0 得x=3,y=1 因为(3-1)²+(1-2)²=5

溪湖区13475318018： 已知直线L:(2m+1)x+(1 - m)y - 3=0. 求证:不论m取何值 L恒过一 - ？
韩购帕朱： 当m=0得x+y-3=0,当m=1得3x-3=0 由此两方程得x=1,y=2 把M(1,2)代入原方程检验 左边=(2m+1)+2(1-m)-3=0=右边 所以不论m取何值 L恒过一定点M (1,2)2,设直线是a(y-2)=(x-1) 根据点到直线的距离公式/a(0-2)-(2-1)/÷根号(a^2+(-1)^2)=1 所以a=0或者a=-4/3 直线L1的方程是x=1或者3x+4y-11=0

溪湖区13475318018： 已知 圆C:(x - 1)^2+(y - 2)^2=25,直线L:(2m+1)x+(m+1)y - 7m - 4=0？
韩购帕朱： (1)直线L:(2m+1)x+(m+1)y-7m-4=0 可化为(2x+y-7)m+(x+y-4)=0 令2x+y-7=0,x+y-4=0 解得x=3,y=1 所以直线恒过定点(3,1) (2)当直线过圆点时 直线L被圆C截得的弦何时最长 把圆点(1,2)代入直线方程得 m=-1/3 当直线垂直于由点(1,2)...

溪湖区13475318018： 已知圆c;(x - 1)平方+(y - 2)平方=25及直线L:(2m+1)x+(m+1)y=7m+4(m - ？
韩购帕朱： 直线L:(2m+1)x+(m+1)y-7m-4=0 直线L:m(2x+y-7)+(x+y-4)=0 令2x+y-7=0;x+y-4=0,解得:x=3;y=1 直线L恒过定点(3,1)