Question:
The equation of a circle is \(\mathrm{Re}(z^2)+2(\mathrm{Im}(z))^2+2\,\mathrm{Re}(z)=0\), 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 y-intercept equal to ______________.
The equation of a circle is \(\mathrm{Re}(z^2)+2(\mathrm{Im}(z))^2+2\,\mathrm{Re}(z)=0\), 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 y-intercept equal to ______________.
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Convert the circle to \(x,y\) form and complete the square for the parabola's vertex.
Updated On: Nov 4, 2026
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Correct Answer: 1
Solution and Explanation
\(\mathrm{Re}(z^2)=x^2-y^2\), so the circle is \(x^2-y^2+2y^2+2x=0\Rightarrow x^2+y^2+2x=0\); centre \((-1,0)\).
Parabola: \(x^2-6x+9=y-4\Rightarrow(x-3)^2=y-4\); vertex \((3,4)\).
Line through \((-1,0)\) and \((3,4)\): slope \(=\frac{4-0}{3+1}=1\), so \(y=x+1\).
y-intercept (put \(x=0\)) \(=1\).
Answer: 1
Parabola: \(x^2-6x+9=y-4\Rightarrow(x-3)^2=y-4\); vertex \((3,4)\).
Line through \((-1,0)\) and \((3,4)\): slope \(=\frac{4-0}{3+1}=1\), so \(y=x+1\).
y-intercept (put \(x=0\)) \(=1\).
Answer: 1
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