第1个回答 2009-09-28
(1)
原心C(-2,0),半径R=2
L1,L2都和圆C相切,且直线L1,L2相互垂直
则:AC=(根号2)R
|a+2|=2(根号2)
a=-2+2(根号2), 或-2-2(根号2)
显然,L1,L2的斜率=1,-1
当a=-2+2(根号2),
直线L1,L2的方程:
y=x+2-2(根号2),
y=-x-2+2(根号2),
当a=-2-2(根号2),
直线L1,L2的方程:
y=x+2+2(根号2),
y=-x-2-2(根号2),
(2)
AM^2=(a-1)^2+m^2=m^2+1
圆M半径^2=AM^2/2=(m^2+1)/2
CM^2=(1+2)^2+m^2=M^2+9
CM=2+((m^2+1)/2)^(1/2)
M^2+9=(2+((m^2+1)/2)^(1/2))^2=4+((m^2+1)/2)+4+((m^2+1)/2)^(1/2)
(M^2+9)^2=32(M^2+1)
M^4-14M^2+49=0
M^2=7, M=+ -(根号7)
圆M的方程: (x-1)^2+(y-m)^2=(m^2+1)/2
(x-1)^2+(y-(根号7))^2=4
或:(x-1)^2+(y+(根号7))^2=4
(3)
设C到l1,l2的距离:p, q, 弦长之和X
则: p^2+q^2=AC^2=1
X=l1弦长 + l2弦长 =2*(R^2-p^2)^(1/2)+2*(R^2-q^2)^(1/2)
=2((4-p^2)^(1/2)+(4-q^2)^(1/2))
(X/2)^2=8-(p^2+q^2)+2((4-p^2)(4-q^2))^(1/2)
=7+2((4-p^2)(4-q^2))^(1/2)
而:(4-p^2)+(4-q^2)=8-(p^2+q^2)=7=定值
当: 4-p^2=4-q^2=7/2时,
(X/2)^2有最大值=7+2*(7/2)=14
X/2=根号14
X最大=2(根号14)
此即所截得弦长之和的最大值