已知实数x,y满足关系x²+y²-2x+4y-20=0,则x²+y²的最小值

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已知实数x,y满足关系x²+y²-2x+4y-20=0,则x²+y²的最小值
已知实数x,y满足关系x²+y²-2x+4y-20=0,则x²+y²的最小值

已知实数x,y满足关系x²+y²-2x+4y-20=0,则x²+y²的最小值
x²+y²-2x+4y-20=0
(x-1)²+(y+2)²=25
设x-1=cosa ;y+2=sina
x=cosa+1 ;y=sina-2
x²+y²
=(cosa+1)²+(sina-2)²
=cos²a+2cosa+1+sin²a-4sina+4
=1+2√5sin(a+θ)+5
=6+2√5sin(a+θ)
所以最小值为6-2√5

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