已知x²+y²-2x-6y+10=0,求4(x²+y)(x²-y)-(2x²-y)²

已知x²+y²-2x-6y+10=0,求4(x²+y)(x²-y)-(2x²-y)²
已知x²+y²-2x-6y+10=0,求4(x²+y)(x²-y)-(2x²-y)²

已知x²+y²-2x-6y+10=0,求4(x²+y)(x²-y)-(2x²-y)²
即(x²-2x+1)+(y²-6y+9)=0
(x-1)²+(y-3)²=0
所以x-1=0,y-3=0
x=1,y=3
所以原式=4x⁴-4y²-4x⁴+4x²y-y²
=4x²y-5y²
=4*1*3-5*9
=-33