$${V_r} = \frac{1}{r}\frac{{\partial \psi }}{{\partial \theta }} = \left( {{V_\infty }\cos \theta } \right)\left( {1 – \frac{{{R^2}}}{{{r^2}}}} \right)$$ and  $${V_\theta } =  – \frac{{\partial \psi }}{{\partial r}} =  – \left( {1 + \frac{{{R^2}}}{{{r^2}}}} \right){V_\infty }\sin \theta$$ therefore,  $$\frac{{{V_r}}}{{{V_\infty }}} = \left( {1 – \frac{{{R^2}}}{{{r^2}}}} \right)\cos \theta$$ $$\frac{{{V_\theta }}}{{{V_\infty }}} =  – \left( {1 + \frac{{{R^2}}}{{{r^2}}}} \right)\sin \theta$$ So, at any given point $\left( {r,\theta } \right)$, ${V_r}$ and ${{V_\theta }}$ are both directly proportional to${{V_\infty }}$.Therefore, the direction of the resultant,$\mathop V\limits^ \to$ is the same, for any value of ${{V_\infty }}$.This infers that the shape of the streamlines remains the same.