Let $\alpha = x+iy$ and write the given circle conditions in complex form.

Since $\alpha$ lies on $(x-x_0)^2+(y-y_0)^2=r^2,$ we have $|\alpha-z_0|=r.$

Also, $\dfrac{1}{\overline{\alpha}}$ lies on $(x-x_0)^2+(y-y_0)^2=4r^2,$ so $\left|\frac{1}{\overline{\alpha}}-z_0\right|=2r.$

Rewrite the second condition.

Note that $\frac{1}{\overline{\alpha}}=\frac{\alpha}{|\alpha|^2}.$ Hence

Square both given relations.

From $|\alpha-z_0|=r$, $|\alpha|^2+|z_0|^2-2\Re(\alpha\overline{z_0})=r^2. \qquad (1)$

From $|1-z_0\overline{\alpha}|=2r|\alpha|$, $1+|z_0|^2|\alpha|^2-2\Re(z_0\overline{\alpha})=4r^2|\alpha|^2. \qquad (2)$

Since $\Re(z_0\overline{\alpha})=\Re(\alpha\overline{z_0})$, let $t=\Re(\alpha\overline{z_0}).$

Then (1) becomes $|\alpha|^2+|z_0|^2-2t=r^2.$ So $2t=|\alpha|^2+|z_0|^2-r^2.$

Substitute into (2).

Equation (2) becomes $1+|z_0|^2|\alpha|^2-2t=4r^2|\alpha|^2.$ Replacing $2t$, $1+|z_0|^2|\alpha|^2-(|\alpha|^2+|z_0|^2-r^2)=4r^2|\alpha|^2.$

Rearranging, $1-|\alpha|^2-|z_0|^2+r^2+|z_0|^2|\alpha|^2=4r^2|\alpha|^2.$

Thus $(|z_0|^2-4r^2-1)|\alpha|^2+(1-|z_0|^2+r^2)=0.$

Use the given relation between $|z_0|^2$ and $r^2$.

Given $2|z_0|^2=r^2+2 \quad\Rightarrow\quad r^2=2|z_0|^2-2.$

Let $u=|z_0|^2$. Then $r^2=2u-2$.

Substitute into the coefficients:

$|z_0|^2-4r^2-1=u-4(2u-2)-1=7-7u=7(1-u),$ and $1-|z_0|^2+r^2=1-u+(2u-2)=u-1=-(1-u).$

Therefore, $7(1-u)|\alpha|^2-(1-u)=0.$

So $(1-u)(7|\alpha|^2-1)=0.$

Determine the valid case.

If $u=1$, then $|z_0|^2=1$, and from $r^2=2u-2=0,$ the circle radius becomes zero, which is not a proper circle here. Hence we take $7|\alpha|^2-1=0.$

Therefore, $|\alpha|^2=\frac{1}{7} \quad\Rightarrow\quad |\alpha|=\frac{1}{\sqrt{7}}.$

Compare with options.

Correct option is $\boxed{\frac{1}{\sqrt7}}.$ This is option $\boxed{\text{C}}$.