If A (cosa, sina), B(sina,-cosa),C(1,2) are the vertices of the triangle ABC, then as 'a' varies , the locus of its centroid is

let x,y be the centroid of the triangle

x=(cos a +sin a +1)/3   which implies 3x-1 = cos a +sin a............(1)

y=(sin a - cos a +2 )/3 which implies 3y-2 = sin a - cos a.............(2)

from here we get, sin a = (3x +3y -3)/2

cos a = (3x -3y + 1)/2

use sin^2 a + cos^2 a =1

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

which is the equation of the circle.

alternate:

squaring and adding (1) and (2)

(3x-1)^2 + (3y-2)^2=2 which is again a circle