Let $\left|\dfrac{\bar z-i}{2\bar z+i}\right|=\dfrac{1}{3},\ z\in\mathbb{C}$, be the equation of a circle with center at $C$. If the area of the triangle whose vertices are at the points $(0,0),\ C$ and $(\alpha,0)$ is $11$ square units, then $\alpha^{2}$ equals:
Step-by-Step Solution
Key Concept: $\left|\dfrac{\bar z-i}{2\bar z+i}\right|=k$ with $k\ne \tfrac12$ defines an \emph{Apollonius-type} locus, which is a circle. Square the modulus equation, expand, and read off the center from the resulting $x^{2}+y^{2}+\dots$ form.
Let $z=x+iy$, $\bar z=x-iy$. Then $\bar z-i = x-i(y+1)$ and $2\bar z+i = 2x-i(2y-1)$.
The condition $9|\bar z-i|^{2}=|2\bar z+i|^{2}$ gives
$$9\bigl(x^{2}+(y+1)^{2}\bigr) = 4x^{2}+(2y-1)^{2}.$$
Expand: $9x^{2}+9y^{2}+18y+9 = 4x^{2}+4y^{2}-4y+1$, so
$$5x^{2}+5y^{2}+22y+8=0\ \Longrightarrow\ x^{2}+y^{2}+\frac{22}{5}y+\frac{8}{5}=0.$$
Center $C=\left(0,-\dfrac{11}{5}\right)$.
Area of triangle with vertices $(0,0),\ \left(0,-\tfrac{11}{5}\right),\ (\alpha,0)$:
$$\text{Area} = \tfrac{1}{2}\,|\alpha|\cdot\tfrac{11}{5} = 11\ \Longrightarrow\ |\alpha|=10\ \Longrightarrow\ \alpha^{2}=100.$$
Correct Answer: 2