Question

The locus of the centroid of the triangle whose vertices are $\left(3 cos \alpha , 3 sin \alpha \right)$ , $\left(9 sin \alpha , - 9 cos \alpha \right)$ and $\left(1 , 0\right)$ is a circle of radius $R\,$ , then the value of $\frac{9 R^{2}}{2}$ is equal to.

Answer 1

Let centroid of $\Delta ABC$ is $\left(h , k\right)$
$\Rightarrow h=\frac{3 cos \alpha + 9 sin \alpha + 1}{3}$ & $k=\frac{3 sin \alpha - 9 cos \alpha + 0}{3}$
$\Rightarrow 3\left(cos \alpha + 3 sin \alpha \right)=3h-1$ & $3\left(sin \alpha - 3 cos \alpha \right)=3k$
$\Rightarrow 9\left(cos \alpha + 3 sin \alpha \right)^{2}=\left(3 h - 1\right)^{2}$ & $9\left(sin \alpha - 3 cos \alpha \right)^{2}=9k^{2}$
$\Rightarrow \left(3 h - 1\right)^{2}+9k^{2}=9\left[\left(cos \alpha + 3 sin \alpha \right)^{2} + \left(sin \alpha - 3 cos \alpha \right)^{2}\right]$
Replace $h$ by $x$ and $k$ by $y$
$\Rightarrow(3 x-1)^{2}+9 y^{2}=90\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)$
$\Rightarrow \left(x - \frac{1}{3}\right)^{2}+y^{2}=\frac{90}{9}$
$\Rightarrow \left(x - \frac{1}{3}\right)^{2}+y^{2}=10$
$\Rightarrow R=\sqrt{10}$
$\Rightarrow \frac{9 R^{2}}{2}=\frac{9 \left(10\right)}{2}=45$