1. Let $\alpha = z$ and define $w=\frac{1}{\bar{\alpha}}.$

   We are given that: $|z-z_0|=2 \quad \text{and} \quad |w-z_0|=4,$ where $z_0=1+i.$

2. A useful identity: $\frac{1}{\bar{z}}=\frac{z}{|z|^2} \quad (z\neq 0).$ So $w=\frac{z}{|z|^2}.$

3. Let $z=x+iy,$ and denote $r^2=|z|^2=x^2+y^2.$ Then $w=\frac{x+iy}{r^2}.$

4. Since $z$ lies on the circle centered at $1+i$ with radius $2$, $|z-(1+i)|^2=4.$ Expanding, $(x-1)^2+(y-1)^2=4$ $x^2+y^2-2x-2y+2=4$ $r^2-2(x+y)=2.$ Hence $x+y=\frac{r^2-2}{2}. \qquad (1)$

5. Now use the second condition: $\left|\frac{z}{r^2}-(1+i)\right|=4.$ Squaring, $\left(\frac{x}{r^2}-1\right)^2+\left(\frac{y}{r^2}-1\right)^2=16.$ Expand: $\frac{x^2+y^2}{r^4}-2\frac{x+y}{r^2}+2=16.$ Since $x^2+y^2=r^2$, $\frac{1}{r^2}-2\frac{x+y}{r^2}+2=16.$ Multiply by $r^2$: $1-2(x+y)+2r^2=16r^2$ $1-2(x+y)=14r^2. \qquad (2)$

6. Substitute (1) into (2): $1-2\left(\frac{r^2-2}{2}\right)=14r^2$ $1-(r^2-2)=14r^2$ $3-r^2=14r^2$ $3=15r^2$ $r^2=\frac{1}{5}.$

   Therefore, $|\alpha|^2=\frac{1}{5}.$

7. Hence, $100|\alpha|^2=100\cdot \frac{1}{5}=20.$

Final Answer

$\boxed{20}$

The derived answer matches the stored correct answer.