For $t\in \left(0,2\pi \right)$, if ABC is an equilateral triangle with vertices $A\left(\sin t,-\cos t\right)$,$B\left(\cos t,\sin t\right)$ and $C\left(a,b\right)$ such that its orthocenter lies on a circle with centre $\left(1,\frac{1}{3}\right)$, then $\left(a^2-b^2\right)=$

Let $P(\mathrm{h},\mathrm{k})=$ orthocenter of triangle ABC

Given $\mathrm{△}\mathrm{ABC}$ is an equilateral

$\therefore \mathrm{H}=\mathrm{I}=\mathrm{S}=\mathrm{G}$

$\therefore P(\mathrm{h},\mathrm{k})=(\frac{\sin \mathrm{t}+\cos \mathrm{t}+\mathrm{a}}{3},\frac{-\cos \mathrm{t}+\sin \mathrm{t}+\mathrm{b}}{3})$

$\Rightarrow \sin t+\cos t=3h-a,\sin t-\cos t=3k-b$

$\therefore (3h-a{)}^2+(3k-b{)}^2=(\sin t+\cos t{)}^2+(\sin t-\cos t{)}^2$

$\therefore$ The locus of P is  ${(x-\frac{a}{3})}^2+{(y-\frac{b}{3})}^2=\frac{2}{9}$

$\therefore$ Given centre  $=(1,\frac{1}{3})=(\frac{\mathrm{a}}{3},\frac{\mathrm{b}}{3})$

$\therefore \mathrm{a}=3,\mathrm{b}=1\Rightarrow {\mathrm{a}}^2-{\mathrm{b}}^2=8$