b
\(F = \frac{{\partial u}}{{\partial r}}\,\hat r = \frac{K}{{{r^3}}}\,\hat r\)

Since particle is moving in circular path

\(\begin{gathered}
  F = \frac{{m{v^2}}}{r} = \frac{K}{{{r^3}}} \Rightarrow m{v^2} = \frac{K}{{{r^2}}} \hfill \\
  \therefore \,K.E. = \frac{1}{2}m{v^2} = \frac{K}{{2{r^2}}} \hfill \\ 
\end{gathered} \)

Total enerygy \(= P.E. + K.E.\)

\( =  - \frac{K}{{2{r^2}}} + \frac{K}{{2{r^2}}} = zero\,\,\left( {\because \,P.E. = \frac{K}{{2{r^2}}}given} \right)\)