= Solution
Interpret the two radial turning values as $r_{\min}=r_0$ and $r_{\max}=4r_0$; the printed maximum/minimum order is reversed. The potential is $V(r)=Ar^2/2$. At either turning point,
$$
E=\frac{L^2}{2mr^2}+\frac A2r^2.
$$
Equating these two energies gives $L^2=mA r_{\min}^2r_{\max}^2$. A radial impulse changes the radial velocity and energy but exerts no torque about the centre, so it preserves $L$. For <radial-impulse circularization of a harmonic central orbit>, the circular orbit satisfies $L^2/(mr_c^3)=Ar_c$. Hence
$$
\boxed{r_c=\sqrt{r_{\min}r_{\max}}=2r_0.}
$$
The original orbit crosses this radius. Cancelling its radial velocity there leaves the tangential speed required for the circular orbit; the impulse is not applied at an original radial turning point.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2015/paper-2-harmonic-circularization.png]
{title=A radial impulse at the geometric-mean radius turns an elliptical harmonic orbit into a circular orbit}
{source=paper-2-harmonic-circularization.py}
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