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=
1
2
4
3
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