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 $k$ is equal to

Question

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 $k$ is equal to, $a^2+b^2$, $a^2+b^2-1$, $a^2+b^2+1$, 0

MARKS App · Accepted Answer

Centroid is given by $x=\frac{x_1+x_2+x_3}{3} \text { and } y=\frac{y_1+y_2+y_3}{3}$ So, $x=\frac{1+a \cos t+b \sin t}{3}$ and $y=\frac{0+a \sin t+(-b \cos t)}{3}$ $\Rightarrow(3 x-1)=a \cos t+b \sin t$ ...(i) and $3 y=a \sin t-b \cos t$ ...(ii) Squaring and adding Eqs. (i) and (ii), we get $\begin{aligned} & 9 x^2-6 x+1+9 y^2=a^2+b^2 \ & \Rightarrow 9 x^2+9 y^2-6 x=a^2+b^2-1 \ & \therefore \quad k=a^2+b^2-1 \ \end{aligned}$

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