若x^2+y^2-2x-2y-3=0,则2x+y-1的最小值是?

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若x^2+y^2-2x-2y-3=0,则2x+y-1的最小值是?
若x^2+y^2-2x-2y-3=0,则2x+y-1的最小值是?

若x^2+y^2-2x-2y-3=0,则2x+y-1的最小值是?
(x-1)^2+(y-1)^2=5
令x-1=√5*cosa
x=1+√5*cosa
则(y-1)^2=5-5(cosa)^2=5(sina)^2
因为sina的值域关于原点对称
所以不妨令y-1=√5*sina
y=1+√5*cosa
2x+y-1=√5*sina+2√5*cosa+2
=√[(√5)^2+(2√5)^2]*sin(a+arctan2)+2
=5sin(a+arctan2)+2
-5

x^2+y^2-2x-2y-3=0
(x^2-2x+1)+(y^2-2y+1)-5=0
(x-1)^2+(y-1)^2=5
所以x-1>=0 y-1>=0
所以x>=1 y>=1
当x和y 都取最小值1时，2x+y-1有最小值
这时2x+y-1=2*1+1-1=2
2x+y-1最小值是2

设2x+y-1=a,y=-2x+1+a
x^2+(-2x+1+a)^2-2x-2(-2x+1+a)-3=0
x^2+4x^2-4x+1+a^2-4ax+2a-2x+4x-2-2a-3=0
5x^2-4ax-2x+a^2-4=0
判别式=(-4a-2)^2-20(a^2-4)=16a^2+16a+4-20a^2+80=-4a^2+16a+84>=0
-3<=a<=7
2x+y-1的最小值是-3

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