From figure, we observe that,
$\begin{gathered}
\theta=\frac{\pi}{4} \\
\therefore \quad A=\left(\cos \frac{\pi}{4}, \sin \frac{\pi}{4}\right)=\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \\
B=\left(\sin \frac{\pi}{4},-\cos \frac{\pi}{4}\right)=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)
\end{gathered}$



and $\quad 0=(0,0)=$ origin
$\therefore$ Centroid of $\triangle A B O$
$=\left\{\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+0 \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+0}{3}, \frac{\sqrt{2}}{3}\right\}$
$=\left(\frac{2}{3 \sqrt{2}}, 0\right)=\left(\frac{\sqrt{2}}{3}, 0\right)$
Let $(h, k)$ be centroid of $\triangle O A B$.
Then, $\quad(h, k)=\left(\frac{\sin \theta+\cos \theta}{3}, \frac{\sin \theta-\cos \theta}{3}\right)$ $\Rightarrow \quad \sin \theta+\cos \theta=3 h \quad \ldots$ (i) and $\quad \sin \theta-\cos \theta=3 k \quad \ldots$ (ii) Now, on adding Eqs. (i) and (ii) after squaring, we get
$\begin{aligned}
&(3 h)^{2}+(3 k)^{2}=(\sin \theta+\cos \theta)^{2} \\
&\quad+(\sin \theta-\cos \theta)^{2}=2 \\
&\Rightarrow \quad h^{2}+k^{2}=\frac{2}{9}
\end{aligned}$
Hence, required locus is
$x^{2}+y^{2}=\frac{2}{9} \text { or } 9 x^{2}+9 y^{2}=2$