= Solution
Hold the two-sphere object fixed in a uniform stream $\mathbf V$. In the <free-draining approximation>, its drag <forces> are $\mathbf f_i=\zeta_i\mathbf V$, with $\zeta_i=6\pi\mu a_i$. The <hydrodynamic torque> about $\mathbf r_c$ is
$$
\mathbf T_c=\sum_{i=1}^2(\mathbf r_i-\mathbf r_c)\times\zeta_i\mathbf V
=\left[\zeta_1\mathbf r_1+\zeta_2\mathbf r_2-(\zeta_1+\zeta_2)\mathbf r_c\right]\times\mathbf V.
$$
To make it zero for every incoming-stream direction, choose the <hydrodynamic centre of a two-sphere dumbbell>:
$$
\boxed{\mathbf r_c=\frac{\zeta_1\mathbf r_1+\zeta_2\mathbf r_2}{\zeta_1+\zeta_2}=\frac{a_1\mathbf r_1+a_2\mathbf r_2}{a_1+a_2}.}
$$
It lies closer to the larger sphere, whose larger drag requires a shorter lever arm. Uniform translation does not spin an isolated sphere, so no additional isolated-sphere spin <torque> enters this leading approximation.
\Image[/past-exam-of-the-advanced-fluid-dynamics-course-of-the-university-of-oxford/2025-two-sphere-hydrodynamic-centre.png]
{title=Drag-weighted hydrodynamic centre of two widely separated spheres}
Finite-separation <hydrodynamic interactions> and connector drag can change the displayed location; the calculation has the same approximation as part <2/c>.
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