The equation of a circle is $\operatorname{Re}\left(z^{2}\right)+2(\operatorname{Im}(z))^{2}+2 \oper

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

Professor PC Thomas Classes · Answer

(1): The equation of circle is\ $\operatorname{Re}\left(z^{2}ight)+2(\operatorname{Im}(z))^{2}+2 \operatorname{Re}(z)=0$\ $\Rightarrow \quad x^{2}-y^{2}+2 y^{2}+2 x=0$\ $\Rightarrow x^{2}+y^{2}+2 x=0 \Rightarrow$ Centre $\equiv(-1,0)$\ Also, $x^{2}-6 x-y+13=0$\ $\Rightarrow \quad(x-3)^{2}=y-13+9$\ $\Rightarrow \quad(x-3)^{2}=y-4 \Rightarrow$ Vertex $\equiv(3,4)$\ $	herefore \quad$ Equation of line passing through $(-1$, 0 ) and ( 3,4 ) is\ $y=\frac{4}{4}(x+1) \Rightarrow y-x=1$\ $	herefore \quad y$-intercept of line $=1$\