A) \[{{x}^{2}}+{{y}^{2}}-2x-4y+1=0\]
B) \[3({{x}^{2}}+{{y}^{2}})-2x-4y+1=0\]
C) \[{{x}^{2}}+{{y}^{2}}-2x-4y+3=0\]
D) None of these
Correct Answer: B
Solution :
Let \[(h,\,\,k)\] be the centroid of the triangle, then \[h=\frac{\cos \alpha +\sin \alpha +1}{3}\] and \[k=\frac{\sin \alpha -\cos \alpha +2}{3}\] \[\Rightarrow \,\,3h-1=\cos \alpha +\sin \alpha \] and \[3k-2=\sin \alpha -\cos \alpha \] \[\Rightarrow \,\,{{(3h-1)}^{2}}+{{(3k-2)}^{2}}=2\], (squaring and adding) \[\Rightarrow \,9\,({{h}^{2}}+{{k}^{2}})-6h-12k+3=0\] \[\Rightarrow \,\,3\,({{h}^{2}}+{{k}^{2}})-2h-4k+1=0\] \[\therefore \] Locus of \[(h,\,\,k)\]is \[3\,({{x}^{2}}+{{y}^{2}})-2x-4y+1=0\].
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