(1)令x+y=z,则y=-x+z,代入圆方程有2x^2-2(z+5)x+z^2+6z+12=0令⊿=[-2(z+5)]^2-4*2*(z^2+6z+12)=0即z^2+2z-1=0,解得z1=-√2-1,z2=√2-1所以(x+y)min=-√2-1,(x+y)max=√2-1(2)令y/x=k,则y=kx,代入圆方程有(1+k^2)x^2+...