Question:
The equation of a circle is
$\operatorname{Re}\left(z^{2}\right)+2(\operatorname{Im}(z))^{2}+2 \operatorname{Re}(z)=0$, where $z=x+i y$.
A line which passes through the center of the given circle and the vertex of the parabola, $x^{2}-6 x-y+13=0$, has $y$-intercept equal to_______.
Solution:
Equation of circle is $\left(x^{2}-y^{2}\right)+2 y^{2}+2 x=0$
$x^{2}+y^{2}+2 x=0$
Centre: $(-1,0)$
Parabola : $x^{2}-6 x-y+13=0$
$(x-3)^{2}=y-4$
Vertex : $(3,4)$
Equation of line $\equiv y-0=\frac{4-0}{3+1}(x+1)$
$y=x+1$
$y$-intercept $=1$
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