= Solution
Use the <angular-diameter ratio under a collinear boost>, with motion along the common sightline and small angular radius $\vartheta$. For a boost toward the source, <relativistic aberration> of the incoming light gives $\tan(\vartheta'/2)=e^{-\eta}\tan(\vartheta/2)$, where $\eta$ is the relative <rapidity>. Thus the observer seeing the smaller disk has <angular diameter> reduced by $e^{-\eta}$. The ratio two implies $e^\eta=2$, so
$$
\boxed{\frac{v_{\rm rel}}c=\tanh(\ln2)=\frac{2^2-1}{2^2+1}=\frac35}.
$$
The answer is \b[$0.60c$]. Both observations occur at one event, so this is a <relativistic aberration> comparison rather than a change of geometrical distance during observation. Arbitrary transverse relative motion would require additional directional information; the stated motion toward the same distant target selects the collinear interpretation.
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