The equation of a circle is operatorname{Re}( | vedclass

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(A) Given the equation of the circle: $\operatorname{Re}(z^{2})+2(\operatorname{Im}(z))^{2}+2 \operatorname{Re}(z)=0$.Since $z=x+iy$,we have $z^{2}=x^

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$1$

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(A) Given the equation of the circle: $\operatorname{Re}(z^{2})+2(\operatorname{Im}(z))^{2}+2 \operatorname{Re}(z)=0$.Since $z=x+iy$,we have $z^{2}=x^{2}-y^{2}+2ixy$,so $\operatorname{Re}(z^{2})=x^{2}-y^{2}$ and $\operatorname{Im}(z)=y$.The equation becomes $(x^{2}-y^{2})+2y^{2}+2x=0$,which simplifies to $x^{2}+y^{2}+2x=0$.The center of this circle is $(-1, 0)$.Given the parabola: $x^{2}-6x-y+13=0$.Rewriting as $(x-3)^{2}-9-y+13=0$,we get $(x-3)^{2}=y-4$.The vertex of the parabola is $(3, 4)$.The line passes through $(-1, 0)$ and $(3, 4)$.The slope $m = \frac{4-0}{3-(-1)} = \frac{4}{4} = 1$.The equation of the line is $y-0=1(x+1)$,which is $y=x+1$.The $y$-intercept is the value of $y$ when $x=0$,which is $1$.

The equation of a circle is $\operatorname{Re}(z^{2})+2(\operatorname{Im}(z))^{2}+2 \operatorname{Re}(z)=0$,where $z=x+iy$. $A$ line which passes through the center of the given circle and the vertex of the parabola $x^{2}-6x-y+13=0$ has a $y$-intercept equal to $.....$

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