(x-2)^2+(y-2)^2=1 ,x^2+Y^2-x-y+12=0 这个方程组解出来是多少

发布时间:2021-03-09 20:00:37

(x-2)^2+(y-2)^2=1 ,x^2+Y^2-x-y+12=0 这个方程组解出来是多少

网友回答

(x-2)^2+(y-2)^2=1
x^2+Y^2-x-y+12=0
x^2+Y^2-4x-4y+8=1
x^2+Y^2-x-y+12=0
x^2+Y^2-x-y+12=0
x^2-x+Y^2-y+12=0
x^2-x+1/4-1/4+Y^2-y+1/4-1/4+12=0
(x-1/2)^2-1/4+(y-1/2)^2-1/4+12=0
(x-1/2)^2+(y-1/2)^2+12-1/4-1/4=0
(x-1/2)^2+(y-1/2)^2+23/2=0
(x-1/2)^2+(y-1/2)^2+23/2=0
(x-1/2)^2+(y-1/2)^2+23/2>0题意矛盾该方程组无解
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