x²+4x+y²-6y+13=0
(x^2+4x+4)+(y^2-6y+9)=0
(x+2)^2+(y-3)^2=0
因为(x+2)^2>=0 (y-3)^2>=0
所以(x+2)^2=0 (y-3)^2=0
x+2=0 y-3=0
x=-2 y=3
x^2+y^2=(-2)^2+3^2=4+9=13
已知x²+4x+y²-6y+13=0,求x²+y²的值
x²+4x+y²-6y+13=0
(x^2+4x+4)+(y^2-6y+9)=0
(x+2)^2+(y-3)^2=0
因为(x+2)^2>=0 (y-3)^2>=0
所以(x+2)^2=0 (y-3)^2=0
x+2=0 y-3=0
x=-2 y=3
x^2+y^2=(-2)^2+3^2=4+9=13