Let a=1+i and z=x+iy. If the curve zbar{z}+az | vedclass

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(C) Given the circle equation $z\bar{z}+az+\bar{a}\bar{z}-4=0$. Since $z\bar{z}=|z|^2=x^2+y^2$ and $a=1+i$, we have $az=(1+i)(x+iy)=(x-y)+i(x+y)$.Thus

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$x^2+y^2+6x+2y=0$

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(C) Given the circle equation $z\bar{z}+az+\bar{a}\bar{z}-4=0$. Since $z\bar{z}=|z|^2=x^2+y^2$ and $a=1+i$, we have $az=(1+i)(x+iy)=(x-y)+i(x+y)$.Thus, $az+\bar{a}\bar{z}=2\operatorname{Re}(az)=2(x-y)$.The equation of the circle becomes $x^2+y^2+2(x-y)-4=0$, which is $S: x^2+y^2+2x-2y-4=0$.The line equation is $(z+\bar{z})-i(z-\bar{z})+2=0$. Since $z+\bar{z}=2x$ and $z-\bar{z}=2iy$, we have $2x-i(2iy)+2=0$, which simplifies to $2x+2y+2=0$ or $L: x+y+1=0$.The equation of the family of circles passing through the intersection of $S$ and $L$ is $S+\lambda L=0$, i.e., $(x^2+y^2+2x-2y-4)+\lambda(x+y+1)=0$.Since the circle passes through the origin $(0,0)$, we substitute $x=0, y=0$ into the equation: $-4+\lambda(1)=0$, which gives $\lambda=4$.Substituting $\lambda=4$ back into the family equation: $x^2+y^2+2x-2y-4+4(x+y+1)=0$.Simplifying, we get $x^2+y^2+6x+2y=0$.

Let $a=1+i$ and $z=x+iy$. If the curve $z\bar{z}+az+\bar{a}\bar{z}-4=0$ is cut by the straight line $(z+\bar{z})-i(z-\bar{z})+2=0$ at two points $A$ and $B$, then the equation of the circle passing through the origin, $A$ and $B$ is

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