已知直线l:kx-y-3k=0,圆M:x^2+y^2-8x-2y+9=0.求证线与圆必相交当圆M截直线l所得弦长最短时,求k的值,并求直线l的方程
解ï¼
L:kx-y-3k=0
y=k*(x-3),Lå¿ è¿ç¹A(3,0)
åM:x^2+y^2-8x-2y+9=0
(x-4)^2+(y-1)^2=8,åå¿M(4,1),r=2â2
MA=â2<r,Aå¨åMå ,æ ç´çº¿Lä¸åMå¿ ç¸äº¤
设y=kx-3kä¸åM交äºB(x1,y1),C(x2,y2),解ä¸æ¹ç¨ç»:
y=kx-3k......(1)
x^2+y^2-8x-2y+9=0......(2)
x^2+(kx-3k)^2-8x-2*(kx-3k)+9=0
(1+k^2)*x^2-(8+2k+6k^2)+9+6k+9k^2=0
å¾x1,2=[(8+2k+6k^2)±2â(7+2k+7k^2)]/[2*(1+k^2)]
设x1>x2,å
x1-x2=2â(7+2k+7k^2)/(1+k^2)
y1-y2=k*(x1-x2)
弦é¿=h
h^2=(x1-x2)^2+(y1-y2)^2
=(x1-x2)^2+[k(x1-x2)]^2
=(1+k^2)*(x1-x2)^2
=4(1+k^2)*(7+2k+7k^2)/(1+k^2)^2
=4(7+2k+7k^2)/(1+k^2)
=4*[7*(1+k^2)+2k]/(1+k^2)
=28+8k/(1+k^2)
设8k/(1+k^2)=n
8k=n+n*k^2
n*k^2-8k+n=0
æªç¥æ°ä¸ºkçä¸æ¹ç¨,æå®æ°è§£çæ¡ä»¶,æ¯å®çå¤å«å¼â¥0,å³
64-4n^2â¥0
n^2â¤16
-4â¤nâ¤4
næå°=-4
å¯å¾h^2æå°=28-4=24,弦æå°=2â6
n=-4
n*k^2-8k+n=0
-4k^2-8k-4=0
k=-1
å½åMæªç´çº¿Læå¾å¼¦é¿æå°æ¶,k=-1 æ以ç´çº¿lçæ¹ç¨-x-y+3=0ä¹å¯ä»¥åæx+y-3=0
L:kx-y-3k=0
y=k*(x-3),Lå¿ è¿ç¹A(3,0)
åM:x^2+y^2-8x-2y+9=0
(x-4)^2+(y-1)^2=8,åå¿M(4,1),r=2â2
MA=â2<r,Aå¨åMå ,æ ç´çº¿Lä¸åMå¿ ç¸äº¤
设y=kx-3kä¸åM交äºB(x1,y1),C(x2,y2),解ä¸æ¹ç¨ç»:
y=kx-3k......(1)
x^2+y^2-8x-2y+9=0......(2)
x^2+(kx-3k)^2-8x-2*(kx-3k)+9=0
(1+k^2)*x^2-(8+2k+6k^2)+9+6k+9k^2=0
å¾x1,2=[(8+2k+6k^2)±2â(7+2k+7k^2)]/[2*(1+k^2)]
设x1>x2,å
x1-x2=2â(7+2k+7k^2)/(1+k^2)
y1-y2=k*(x1-x2)
弦é¿=h
h^2=(x1-x2)^2+(y1-y2)^2
=(x1-x2)^2+[k(x1-x2)]^2
=(1+k^2)*(x1-x2)^2
=4(1+k^2)*(7+2k+7k^2)/(1+k^2)^2
=4(7+2k+7k^2)/(1+k^2)
=4*[7*(1+k^2)+2k]/(1+k^2)
=28+8k/(1+k^2)
设8k/(1+k^2)=n
8k=n+n*k^2
n*k^2-8k+n=0
æªç¥æ°ä¸ºkçä¸æ¹ç¨,æå®æ°è§£çæ¡ä»¶,æ¯å®çå¤å«å¼â¥0,å³
64-4n^2â¥0
n^2â¤16
-4â¤nâ¤4
næå°=-4
å¯å¾h^2æå°=28-4=24,弦æå°=2â6
n=-4
n*k^2-8k+n=0
-4k^2-8k-4=0
k=-1
å½åMæªç´çº¿Læå¾å¼¦é¿æå°æ¶,k=-1 æ以ç´çº¿lçæ¹ç¨-x-y+3=0ä¹å¯ä»¥åæx+y-3=0
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