问题
填空题
若直线ax+2by-2=0(a>0,b>0)始终平分圆x2+y2-4x-2y-8=0的周长,则
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答案
x2+y2-4x-2y-8=0可化为:(x-2)2+(y-1)2=13,∴圆的圆心是(2,1)
∵直线平分圆的周长,所以直线恒过圆心(2,1)
把(2,1)代入直线ax+2by-2=0,得a+b=1
∴
+1 a
=(2 b
+1 a
)(a+b)=3+2 b
+b a 2a b
∵a>0,b>0,
∴
+1 a
=(2 b
+1 a
)(a+b)=3+2 b
+b a
≥3+22a b 2
0≤ab≤(
)2=a+b 2 1 4
故答案为:3+2
,(0,2
]1 4