Question

If $A(\cos \alpha ,\sin \alpha ),B(\sin \alpha ,-\cos \alpha ),C(1,2)$ are the vertices of $\Delta ABC$, then as $\alpha$ varies, the locus of its centroid is

• A. $x^2+y^2-2x-4y+3=0$
• B. $x^2+y^2-2x-4y+1=0$
• C. $3(x^2+y^2)-2x-4y+1=0$
• D. $Noneofthese$

Answer 1

The correct option is C $3(x^2+y^2)-2x-4y+1=0$

Let $G(h,k)$ be the centroid of the $△ABC$
$\therefore (h,k)=(\frac{\cos \alpha +\sin \alpha +1}{3},\frac{\sin \alpha -\cos \alpha +2}{3})$
$\Rightarrow \cos \alpha +\sin \alpha =3h-1.......\left(1\right)$
and $\sin \alpha -\cos \alpha =3k-2......\left(2\right)$
Now squaring and adding $\left(1\right)$ and $\left(2\right)$ we get,
$2=(3h-1{)}^2+(3k-2{)}^2$
$\Rightarrow 9h^2+9k^2-6h-12k+3=0$
$\Rightarrow 3(h^2+k^2)-2h-4k+1=0$
Hence required locus is, $3(x^2+y^2)-2x-4y+1=0$