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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 trianglex=(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)/2cos a = (3x -3y + 1)/2use sin^2 a + cos^2 a =1(3x + 3y-3)^2 + (3x - 3y+1)^2 = 4which 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
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
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