(1)设直线BC为y=kx+1,代入x 2+y 2=4得,(1+k 2)x 2+2kx-3=0,
S=
1
2 |FA|| x 1 - x 2 |
=
| x 1 - x 2 |
2
=
( x 1 + x 2 ) 2 -4 x 1 x 2
2
=
4 k 2 +3
(1+ k 2 ) 2
=
4-(
1
1+ k 2 -2 ) 2 ≤
3 .
当且仅当k=0时,△ABC的最大面积为
3 .
(2)设圆心 (a,
a 2
4 ) ,则圆为 (x-a ) 2 +(y-
a 2
4 ) 2 = a 2 +(2-
a 2
4 ) 2 .
当y=0时,x=a±2,
∴|MN|=4,
令∠MAN=θ,
由余弦定理,得16=m 2+n 2-2mncosθ,
又由 S △AMN =
1
2 mnsinθ-
1
2 |MN| y A
=
1
2 ×4×2=4 ,
∴
16
mn =2sinθ ,
∴
m
n +
n
m =2(sinθ +cosθ+
=2
2 sin(θ+
π
4 ) ≤2
2 ,
当 θ=
π
4 时取得最大值.