Complex Numbers
Apollonius Circle — Centre and Area
nta_pyq_2025_apr
Grade 11
Question:
Let $\left|\dfrac{z-i}{2z+i}\right|=\dfrac{1}{3}$, $z\in\mathbb{C}$, be the equation of a circle with centre $C$. If the area of the triangle whose vertices are at $(0,0)$, $C$, and $(\alpha,0)$ is 11 square units, then $\alpha^2$ equals
50
100
$\dfrac{81}{25}$
$\dfrac{121}{25}$
Step-by-Step Solution
Key Concept: Square both sides to get an Apollonius circle equation, complete the square to find the centre, then use area $=\tfrac{1}{2}\cdot|\alpha|\cdot|y_C|=11$.
$|z-i|^2=\tfrac{1}{9}|2z+i|^2$ with $z=x+iy$:
$x^2+(y-1)^2=\tfrac{1}{9}\left[4x^2+(2y+1)^2\right]$
$9x^2+9y^2-18y+9=4x^2+4y^2+4y+1$
$5x^2+5y^2-22y+8=0 \Rightarrow x^2+\left(y-\tfrac{11}{5}\right)^2=\dfrac{121}{25}-\dfrac{8}{5}=\dfrac{81}{25}$.
Centre $C=\left(0,\dfrac{11}{5}\right)$.
Area $=\dfrac{1}{2}|\alpha|\cdot\dfrac{11}{5}=11 \Rightarrow |\alpha|=10 \Rightarrow \alpha^2=100$.
Correct Answer: 2