Question

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

Answer 1

Given: $Re\left(z^2\right)+2\left(Im\right(z){)}^2+2Re(z)=0,$
$\Rightarrow (x^2-y^2)+2\left(y^2\right)+2x=0$
$\Rightarrow x^2+y^2+2x=0$
Centre, $C\equiv (-1,0)$
Also, parabola is $(x-3{)}^2=(y-4)$
Vertex of parabola is $(3,4)$
Now, the equation of line which passes through the points $(-1,0)$ and $(3,4)$ is
$y-0=\frac{4-0}{3+1}(x+1)$
$\Rightarrow y=x+1$
Hence, $y-$ intercept is $1.$