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Solution

Given: $\left|z_1-2z_2\right| = 2\left|\frac{1}{2}-z_1 z_2\right|$
Square both sides: $|z_1-2z_2|^2 = 4\left|\frac{1}{2}-z_1 z_2\right|^2$
Expand: $|z_1|^2 - 2z_1\overline{z_2} - 2\overline{z_1}z_2 + 4|z_2|^2$
$= 4\left(\frac{1}{4} - \frac{1}{2}z_1 z_2 - \frac{1}{2}\overline{z_1 z_2} + |z_1 z_2|^2\right)$
Simplify RHS: $= 1 - 2z_1 z_2 - 2\overline{z_1 z_2} + 4|z_1|^2|z_2|^2$
Compare both sides and simplify: $\Rightarrow |z_1|^2 + 4|z_2|^2 = 1 + 4|z_1|^2|z_2|^2$
Rearrange: $\Rightarrow (1-4|z_2|^2)|z_1|^2 + (4|z_2|^2-1)=0$
Factor: $\Rightarrow (4|z_2|^2-1)(|z_1|^2-1)=0$
So either: $4|z_2|^2-1=0 \Rightarrow |z_2|=\frac{1}{2}$
or $|z_1|^2-1=0 \Rightarrow |z_1|=1$
Thus: either $z_1$ lies on circle radius $1$ or $z_2$ lies on circle radius $\frac{1}{2}$
Final Answer: $ \boxed{\text{Option (3)}} $