Let $\left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3}, z \in C$, be the equation of a circle with center at $C$. If the area of the triangle, whose vertices are at the points $(0,0), \mathrm{C}$ and $(\alpha, 0)$ is 11 square units, then $\alpha^2$ equals:

Question

Aicharya · Accepted Answer

Let $z=x+i y \Rightarrow \bar{z}=x-i y$
$\begin{aligned}
& 3|\bar{z}-i|=1|2 \bar{z}+i| \\ & =3 \mid(x-(y+1) i|=|2 x+i(1-2 y)| \\ & =3 \sqrt{x^2+(y+1)^2}=\sqrt{(2 x)^2+(1-2 y)^2} \\ & =9\left(x^2+y^2+2 y+1\right)=4 x^2+4 y^2-4 y+1 \\ & \Rightarrow 5 x^2+5 y^2+22 y+8=0 \\ & \Rightarrow \text { Centre } \equiv\left(0,-\frac{11}{5}\right)
\end{aligned}$

Area of $\Delta$
$=\frac{1}{2}|\alpha|\left|\frac{-11}{5}\right|=11$
$\begin{aligned} & \Rightarrow \quad|\alpha|=10 \\ & \Rightarrow \quad \alpha^2=100\end{aligned}$

Let $\left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3}, z \in C$, be the equation of a circle with…

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