\(A \equiv(\cos \theta, \sin \theta), B \equiv(\sin \theta,-\cos \theta)\) are two points. The locus of the centroid of \(\triangle O A B\), where \(O\) is the origin, is

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
\(x^{2}+y^{2}=\frac{2}{9} \text { or } 9 x^{2}+9 y^{2}=2\)