= Solution
The <principle of stationary action>, also called Hamilton's principle, requires $\delta\int_{t_1}^{t_2}L\,dt=0$ for fixed-endpoint path variations. Here the original PDF has $\dot r^2$ in the radial kinetic energy; the TeX has lost that dot. With $m>0$ and $r>0$, the <Lagrangian> is
$$
L=\frac m2\dot r^2+\frac m2r^2\dot\phi^2-kmr^2.
$$
The <Euler-Lagrange equation> for each coordinate gives
$$
\boxed{\ddot r=r\dot\phi^2-2kr,\qquad\frac d{dt}(mr^2\dot\phi)=0.}
$$
The two conserved quantities are <angular momentum> $h=mr^2\dot\phi$ and energy $E=\frac m2\dot r^2+\frac m2r^2\dot\phi^2+kmr^2$. By <Noether's theorem>, they arise respectively from rotational invariance and time-translation invariance of the action. The coordinate $\phi$ is a <cyclic coordinate>, and $L$ has no explicit time dependence.
The <conjugate momentum> for each coordinate is $p_r=m\dot r$, $p_\phi=mr^2\dot\phi$. The <Legendre transform> $H=p_r\dot r+p_\phi\dot\phi-L$ gives the <Hamiltonian>
$$
\boxed{H=\frac{p_r^2}{2m}+\frac{p_\phi^2}{2mr^2}+kmr^2.}
$$
<Hamilton's equations> are
$$
\dot r=\frac{p_r}{m},\quad \dot\phi=\frac{p_\phi}{mr^2},\quad
\dot p_r=\frac{p_\phi^2}{mr^3}-2kmr,\quad \dot p_\phi=0.
$$
Setting $p_\phi=h$ yields $m\ddot r=-V_{\rm eff}'(r)$ with the <effective potential> $V_{\rm eff}=h^2/(2mr^2)+kmr^2$.
For $k>0$ and $h\ne0$, this potential tends to infinity both as $r\downarrow0$ and as $r\to\infty$, and has its unique stationary point at
$$
\boxed{r_0=\left(\frac{h^2}{2km^2}\right)^{1/4},\qquad V_{\rm eff}''(r_0)=8km>0.}
$$
The <effective potential stability criterion> proves a stable <circular orbit in a quadratic central potential>, with $|\dot\phi|=\sqrt{2k}$. For a small radial displacement $\eta$, $\ddot\eta+8k\eta=0$ to first order, so radial perturbations oscillate rather than grow. The diagram shows the centrifugal and quadratic contributions and their sum in dimensionless units:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ib/paper-2-effective-potential.png]
{title=The <effective potential> for a central-force orbit}
The printed paper only says that $k$ is constant. Its claimed positive-radius stable circular orbit needs the additional attractive-force assumption $k>0$ and nonzero angular momentum. For $k<0$ there is no such minimum; for $k=0,h\ne0$ the effective potential is strictly decreasing. With $h=0,k>0$, the minimum is at the origin and is not a positive-radius circular orbit. These cases are not represented by the diagram.
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