已知圆C:x²+(y-1)²=5,直线L:mx-y+1-m=0,设该直线与圆相交于AB两点,若|AB|=√17,求L的倾斜再求直线L中,截圆所得的弦最长及最短时的直线方程.
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![已知圆C:x²+(y-1)²=5,直线L:mx-y+1-m=0,设该直线与圆相交于AB两点,若|AB|=√17,求L的倾斜再求直线L中,截圆所得的弦最长及最短时的直线方程.](/uploads/image/z/1765135-55-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86C%3Ax%26%23178%3B%2B%EF%BC%88y-1%29%26%23178%3B%3D5%2C%E7%9B%B4%E7%BA%BFL%EF%BC%9Amx-y%2B1-m%3D0%2C%E8%AE%BE%E8%AF%A5%E7%9B%B4%E7%BA%BF%E4%B8%8E%E5%9C%86%E7%9B%B8%E4%BA%A4%E4%BA%8EAB%E4%B8%A4%E7%82%B9%2C%E8%8B%A5%7CAB%7C%3D%E2%88%9A17%2C%E6%B1%82L%E7%9A%84%E5%80%BE%E6%96%9C%E5%86%8D%E6%B1%82%E7%9B%B4%E7%BA%BFL%E4%B8%AD%2C%E6%88%AA%E5%9C%86%E6%89%80%E5%BE%97%E7%9A%84%E5%BC%A6%E6%9C%80%E9%95%BF%E5%8F%8A%E6%9C%80%E7%9F%AD%E6%97%B6%E7%9A%84%E7%9B%B4%E7%BA%BF%E6%96%B9%E7%A8%8B.)
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已知圆C:x²+(y-1)²=5,直线L:mx-y+1-m=0,设该直线与圆相交于AB两点,若|AB|=√17,求L的倾斜再求直线L中,截圆所得的弦最长及最短时的直线方程.
已知圆C:x²+(y-1)²=5,直线L:mx-y+1-m=0,设该直线与圆相交于AB两点,若|AB|=√17,求L的倾斜
再求直线L中,截圆所得的弦最长及最短时的直线方程.
已知圆C:x²+(y-1)²=5,直线L:mx-y+1-m=0,设该直线与圆相交于AB两点,若|AB|=√17,求L的倾斜再求直线L中,截圆所得的弦最长及最短时的直线方程.
mx-y+1-m=0 =>y=mx+1-m
代入圆方程 =>
x²+(mx-m)²=5 =>
(1+m²)x²-2m²x+m²-5=0 (1)
设两交点为(x1,y1)(x2,y2)
|AB|=根号[(x2-x1)^2+(y2-y1)^2]=根号(1+m²)|x2-x1| (x2,x1为方程1的两个根)
|x2-x1|=根号[(x1+x2)^2-4x1x2]=根号{2m²/(1+m²)]^2-4(m²-5)/(1+m²)}=√17/根号(1+m²)
=>(2m²)^2-4(m²-5)(1+m²)=17(1+m²)
=>3-m²=0 m=根号3,负根号3.(即为斜率)
|AB|=根号[4m^4/(1+m²)-4(m²-5)]=根号[(20+16m²)/(1+m^2)]=根号[16+4/(1+m^2)]
=> m=0 |AB|=20 取最大值
m=无穷大时,|AB|=16 取最小值