If $$z_1, z_2$$ are two distinct complex numbers such that $$\left|\frac{z_1 - 2z_2}{\frac{1}{2} - z_1\bar{z}_2}\right| = 2$$, then

Given $$\left|\frac{z_1 - 2z_2}{\frac{1}{2} - z_1\bar{z}_2}\right| = 2$$.

Multiply numerator and denominator by 2:

$$\left|\frac{2(z_1 - 2z_2)}{1 - 2z_1\bar{z}_2}\right| = 2$$

$$\left|\frac{z_1 - 2z_2}{\frac{1}{2} - z_1\bar{z}_2}\right| = 2$$

Squaring both sides:

$$|z_1 - 2z_2|^2 = 4\left|\frac{1}{2} - z_1\bar{z}_2\right|^2$$

Expanding the left side:

$$|z_1|^2 - 2z_1\overline{2z_2} - 2z_2\bar{z}_1 + 4|z_2|^2 = |z_1|^2 - 2(z_1\bar{z}_2 + \bar{z}_1 z_2) + 4|z_2|^2$$

Wait, let me be more careful. $$|z_1 - 2z_2|^2 = (z_1 - 2z_2)(\bar{z}_1 - 2\bar{z}_2) = |z_1|^2 - 2z_1\bar{z}_2 - 2\bar{z}_1 z_2 + 4|z_2|^2$$.

Right side: $$4\left|\frac{1}{2} - z_1\bar{z}_2\right|^2 = 4\left(\frac{1}{2} - z_1\bar{z}_2\right)\left(\frac{1}{2} - \bar{z}_1 z_2\right)$$

$$= 4\left(\frac{1}{4} - \frac{1}{2}\bar{z}_1 z_2 - \frac{1}{2}z_1\bar{z}_2 + |z_1|^2|z_2|^2\right)$$

$$= 1 - 2\bar{z}_1 z_2 - 2z_1\bar{z}_2 + 4|z_1|^2|z_2|^2$$

Setting LHS = RHS:

$$|z_1|^2 - 2z_1\bar{z}_2 - 2\bar{z}_1 z_2 + 4|z_2|^2 = 1 - 2\bar{z}_1 z_2 - 2z_1\bar{z}_2 + 4|z_1|^2|z_2|^2$$

The $$-2z_1\bar{z}_2 - 2\bar{z}_1 z_2$$ terms cancel from both sides:

$$|z_1|^2 + 4|z_2|^2 = 1 + 4|z_1|^2|z_2|^2$$

Rearranging: $$|z_1|^2 - 1 + 4|z_2|^2 - 4|z_1|^2|z_2|^2 = 0$$

$$(|z_1|^2 - 1) - 4|z_2|^2(|z_1|^2 - 1) = 0$$

$$(|z_1|^2 - 1)(1 - 4|z_2|^2) = 0$$

So either $$|z_1|^2 = 1$$ or $$|z_2|^2 = \frac{1}{4}$$.

This means either $$|z_1| = 1$$ ($$z_1$$ lies on a circle of radius 1) or $$|z_2| = \frac{1}{2}$$ ($$z_2$$ lies on a circle of radius $$\frac{1}{2}$$).

The answer is Option D.