Thank you for reporting, we will resolve it shortly
Q.
$A \equiv (\cos \theta, \sin \theta), B= (\sin \theta, - \cos \theta) $ are two points. The locus of the centroid of $\Delta OAB$, where $'O'$ is the origin is
From figure, we observe that,
$\theta=\frac{\pi}{4} $
$\therefore A=\left(\cos \frac{\pi}{4}, \sin \frac{\pi}{4}\right)=\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) $
$B=\left(\sin \frac{\pi}{4},-\cos \frac{\pi}{4}\right)=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$
and $0=(0,0)=$ origin
$\therefore $ Centroid of $\Delta A B O$
$=\left\{\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+0}{3}, \frac{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+0}{3}\right\}$
$=\left(\frac{2}{3 \sqrt{2}}, 0\right)=\left(\frac{\sqrt{2}}{3}, 0\right)$
Let $(h, k)$ be centroid of $\Delta O A B$.
Then, $ (h, k)=\left(\frac{\sin \theta+\cos \theta}{3}, \frac{\sin \theta-\cos \theta}{3}\right) $
$\Rightarrow \sin \theta+\cos \theta=3\, h \,\,\,\,\ldots( i) $
and $\sin \theta-\cos \theta=3 \,k\,\,\,\,\,\,\,\,...(ii)$
Now, on adding Eqs. (i) and (ii) after squaring, we get
$(3 h)^{2}+(3 k)^{2}=(\sin \theta+\cos \theta)^{2} +(\sin \theta-\cos \theta)^{2}=2 $
$\Rightarrow n^{2}+k^{2}=\frac{2}{9} $
Hence, required locus is
$x^{2}+y^{2}=\frac{2}{9}\,\,\,$ or $\,\,\,9 x^{2}+9 y^{2}=2$
