Explanation

(b)
Given equation of circle is$x^{2} + y^{2} = r^{2}$. Let any point on the circle is $P(r\cos\theta,\ r\sin\theta)$ and let the coordinates of centriod of the triangle be $(\alpha,\ \beta)$
[image:11_image15.emf]
Then, $\alpha = \frac{r + r\cos\theta}{3}$
$\Rightarrow \ \frac{r}{3}\cos\theta = \alpha - \frac{r}{3}$
and $\beta = \frac{r + r\sin\theta}{3}$
$\Rightarrow \ \frac{r}{3}\sin\theta = \beta - \frac{r}{3}$
Now, $\left( \alpha - \frac{r}{3} \right)^{2} + \left( \beta - \frac{r}{3} \right)^{2} = \frac{r^{2}}{9}$
$\therefore$ The locus is $\left( x - \frac{r}{3} \right)^{2} + \left( y - \frac{r}{3} \right)^{2} = \left( \frac{r}{3} \right)^{2}$ which is a circle