1. Rewrite the locus in terms of $z = x+iy$.

Let $z=x+iy \implies \bar z=x-iy.$ Then $\bar z-i=x-i(y+1), \qquad 2\bar z+i=2x+i(1-2y).$

Given $\left|\frac{\bar z-i}{2\bar z+i}\right|=\frac13,$ so $\frac{|\bar z-i|}{|2\bar z+i|}=\frac13$ which gives $3|\bar z-i|=|2\bar z+i|.$

2. Square both sides.

We have $|\bar z-i|^2=x^2+(y+1)^2,$ and $|2\bar z+i|^2=(2x)^2+(1-2y)^2=4x^2+(1-2y)^2.$

Thus $9\bigl(x^2+(y+1)^2\bigr)=4x^2+(1-2y)^2.$

Expand: $9x^2+9(y^2+2y+1)=4x^2+(1-4y+4y^2).$ So $9x^2+9y^2+18y+9=4x^2+4y^2-4y+1.$

Bring all terms to one side: $5x^2+5y^2+22y+8=0.$ Divide by $5$: $x^2+y^2+\frac{22}{5}y+\frac85=0.$

3. Convert to standard circle form.

Complete the square in $y$: $y^2+\frac{22}{5}y=\left(y+\frac{11}{5}\right)^2-\frac{121}{25}.$ Hence $x^2+\left(y+\frac{11}{5}\right)^2-\frac{121}{25}+\frac85=0.$ Now $\frac85=\frac{40}{25},$ so $x^2+\left(y+\frac{11}{5}\right)^2=\frac{81}{25}.$

Therefore the circle has center $C=\left(0,-\frac{11}{5}\right).$

4. Use the area of the triangle.

The triangle has vertices $(0,0)$, $C=(0,-11/5)$, and $(\alpha,0)$.

This is a right triangle with:

• base along the $x$-axis = $|\alpha|$
• height = distance of $C$ from $x$-axis = $\frac{11}{5}$

So area is $\frac12\cdot |\alpha|\cdot \frac{11}{5}=11.$ Thus $\frac{11|\alpha|}{10}=11$ which gives $|\alpha|=10.$ Hence $\alpha^2=100.$

5. Check with options.

$100$ corresponds to Option B.