jmc

Let the equation of the circle be $(x-a{)}^2+(y-b{)}^2=r^2.$ From $xy=1,$ $y=\frac{1}{x}.$ Substituting, we get $$(x-a{)}^2+{\left(\frac{1}{x}-b\right)}^2=r^2.$$Then $$x^2-2ax+a^2+\frac{1}{x^2}-\frac{2b}{x}+b^2=r^2,$$so $$x^4-2a{x}^3+(a^2+b^2-r^2)x^2-2bx+1=0.$$By Vieta's formulas, the product of the roots is 1. Three of the roots are 2, $-5,$ and $\frac{1}{3},$ so the fourth root is $-\frac{3}{10}.$ Therefore, the fourth point is $\boxed{\left(-\frac{3}{10},-\frac{10}{3}\right)}.$