$A B C$ is a variable triangle with fixed vertex $C (1, 2)$ and $A, B$ having the coordinates $(cos t, sin t), (sin t, - cos t)$ respectively where $t$ is a parameter the locus of the centroid of the $△ A B C$ is:

3 {(x}^{2} + y^{2}) - 2 x - 4 y - 1 = 0

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3 {(x}^{2} + y^{2}) - 2 x - 4 y + 1 = 0

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3 {(x}^{2} + y^{2}) + 2 x + 4 y - 1 = 0

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3 {(x}^{2} + y^{2}) + 2 x + 4 y + 1 = 0

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Solution

The correct option is B $3 {(x}^{2} + y^{2}) - 2 x - 4 y + 1 = 0$
Let coordinates of centroid be (h,k)
$h = \frac{1 + cos t + sin t}{3}$
$k = \frac{2 + sin t - cos t}{3}$
$∴ 3 h - 1 = sin t + cos t ⟶ (1)$
$3 k - 2 = sin t - cos t ⟶ (2)$
Squaring both the sides and add (1) and (2)
We get,
${(3 h - 1)}^{2} + {(3 k - 2)}^{2} = 2$
$9 (h^{2} + k^{2}) - 6 h - 12 k + 3 = 0$
$3 (h^{2} + k^{2}) - 2 h - 4 k + 1 = 0$
replace (h,k) $⟶$ (x,y)
$∴ 3 (x^{2} + y^{2}) - 2 x - 4 y + 1 = 0$

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