Let AB be a chord of the circle x^2 + y^2 = r | vedclass

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(B) Let the coordinates of $A$ be $(r, 0)$ and $B$ be $(0, r)$ such that $\angle AOB = 90^\circ$. Let $P$ be a point on the circle given by $(r \cos \

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$A$ circle

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(B) Let the coordinates of $A$ be $(r, 0)$ and $B$ be $(0, r)$ such that $\angle AOB = 90^\circ$. Let $P$ be a point on the circle given by $(r \cos 	heta, r \sin 	heta)$.The centroid $G(\alpha, \beta)$ of $\Delta PAB$ is given by:$\alpha = \frac{r \cos 	heta + r + 0}{3} = \frac{r(\cos 	heta + 1)}{3}$$\beta = \frac{r \sin 	heta + 0 + r}{3} = \frac{r(\sin 	heta + 1)}{3}$Rearranging,we get:$\cos 	heta = \frac{3\alpha}{r} - 1$$\sin 	heta = \frac{3\beta}{r} - 1$Using the identity $\cos^2 	heta + \sin^2 	heta = 1$:$\left(\frac{3\alpha}{r} - 1ight)^2 + \left(\frac{3\beta}{r} - 1ight)^2 = 1$$\frac{9}{r^2} \left(\alpha - \frac{r}{3}ight)^2 + \frac{9}{r^2} \left(\beta - \frac{r}{3}ight)^2 = 1$$\left(\alpha - \frac{r}{3}ight)^2 + \left(\beta - \frac{r}{3}ight)^2 = \left(\frac{r}{3}ight)^2$Thus,the locus is $(x - \frac{r}{3})^2 + (y - \frac{r}{3})^2 = (\frac{r}{3})^2$,which represents a circle.

Let $AB$ be a chord of the circle $x^2 + y^2 = r^2$ subtending a right angle at the centre. Then the locus of the centroid of the $\Delta PAB$ as $P$ moves on the circle is

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