圆C(x-3)^2+(y-4)^2=4,直线L1过定点A(1,0)若L1与圆相交与P,Q两点,线段PQ的中点为M,又L1与L2:x+2y+2=0的交点为N,求证:AM*AN为定值

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圆C(x-3)^2+(y-4)^2=4,直线L1过定点A(1,0)若L1与圆相交与P,Q两点,线段PQ的中点为M,又L1与L2:x+2y+2=0的交点为N,求证:AM*AN为定值
圆C(x-3)^2+(y-4)^2=4,直线L1过定点A(1,0)
若L1与圆相交与P,Q两点,线段PQ的中点为M,又L1与L2:x+2y+2=0的交点为N,求证:AM*AN为定值

圆C(x-3)^2+(y-4)^2=4,直线L1过定点A(1,0)若L1与圆相交与P,Q两点,线段PQ的中点为M,又L1与L2:x+2y+2=0的交点为N,求证:AM*AN为定值
设直线L1方程为y=k(x-1) (1)
与L2:x+2y+2=0的交点为
N((2k-2)/(2k+1),-3k/(2k+1))
AN^2=(x-1)^2+y^2=(9+9k^2)/(2k+1)^2
=9(k^2+1)^2/(2k+1)^2
CM与直线L1方程为y=k(x-1)为
M((k^2+4k+3)/(k^2+1),(4k^2+2k)/(k^2+1))
AM^2==(x-1)^2+y^2
=(4k+2)^2/(k^2+1)^2+(4k^2+2k)^2/(k^2+1)^2
==(4k+2)^2(1+k^2)/(k^2+1)^2
(AN*AM)^==9(1+k^2)/(2k+1)^2×(4k+2)^2×(1+k^2)/(k^2+1)^2
=9×4=36
∴AM*AN为定值 6
这题做得好辛苦呀

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