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$I f A (\cos α, \sin α), B (\sin α, - \cos α), C (1, 2) a r e t h e v e r t i c e s o f a t r i a n g l e A B C t h e n a s α v a r i e s, t h e l o c u s o f t h e c e n t r o i d o f t h e t r i a n g l e i s k (x^{2} + y^{2}) - 2 x - 4 y + 1 = 0 t h e n k =$

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Correct option is B

Eliminating the parameter α, will get the locus of the centroid of the triangleThe centroid of the triangle is cosα+sinα+13,sinα-cosα+23=x,y(3x-1)=cosα+sinα(3y-2)=sinα-cosα(3x-1)2+(3y-2)2=2cos2α+sin2α

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If Acosα,sinα,Bsinα,-cosα,C1,2 are the vertices of a triangle ABC then as α varies, the locus of the centroid of the triangle is kx2+y2-2x-4y+1=0 then k=

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$I f A (\cos α, \sin α), B (\sin α, - \cos α), C (1, 2) a r e t h e v e r t i c e s o f a t r i a n g l e A B C t h e n a s α v a r i e s, t h e l o c u s o f t h e c e n t r o i d o f t h e t r i a n g l e i s k (x^{2} + y^{2}) - 2 x - 4 y + 1 = 0 t h e n k =$