A particle is moving is moving in a circular path of radius r under the action of an attractive potential $U=k/2r^2$ . Its total energy is

$\begin{array}{l}F=-\frac{dU}{dr}=-\frac{d(-K/2r^2)}{dr}=\frac{K}{2}\frac{d{r}^{-2}}{dr} (on differentiating with respect to r)\\ \\ F=\frac{K}{2}(-2r^{-3})=-\frac{K}{r^3}\end{array}$

Since it is performing circular motion FORCE F=$\frac{m{v}^2}{r}=\frac{K}{r^3}$

$\begin{array}{l}m{v}^2=\frac{K}{r^2}\Rightarrow K.E=\frac{1}{2}m{v}^2=\frac{K}{2r^2}\\ given PE=U=\frac{K}{2r^2}\\ Total Energy=KE+PE=-\frac{K}{2r^2}+\frac{K}{2r^2}=zero\\ \end{array}$