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Q.
Locus of the centroid of a triangle whose vertices are $(1,0),(a \cos t, a \sin t),(b \sin t,-b \cos t)$ is $9 x^{2}+9 y^{2}-6 x=k$. Then the value of $k=$
Centroid is given by
$x=\frac{x_{1}+x_{2}+x_{3}}{3}$
and $y=\frac{y_{1}+y_{2}+y_{3}}{3}$
So, $x=\frac{1+a \cos t+b \sin t}{3}$
and $y=\frac{0+a \sin t+(-b \cos t)}{3}$
$\Rightarrow (3 x-1)=a \cos t+b \sin t \ldots$ (i)
and $3 y=a \sin t-b \cos t \ldots$ (2)
Squaring and adding Eqs. (i) and (ii), we get
$9 x^{2}-6 x+1+9 y^{2}=a^{2}+b^{2}$
$\Rightarrow 9 x^{2}+9 y^{2}-6 x=a^{2}+b^{2}-1$
$\therefore k=a^{2}+b^{2}-1$