A) circle
B) straight line
C) ellipse
D) None of these
Correct Answer: A
Solution :
Let\[A({{x}_{1}},{{y}_{1}}),B({{x}_{2}},{{y}_{2}})\]and\[C({{x}_{3}},{{y}_{3}})\]be the vertices of the triangle ABC, and let\[P(\lambda ,k)\]be any points on the locus. Then, \[P{{A}^{2}}+P{{B}^{2}}+P{{C}^{2}}=c\] (constant) \[\Rightarrow \] \[\sum\limits_{i=1}^{3}{{{(h-{{x}_{i}})}^{2}}+{{(k-{{y}_{i}})}^{2}}=c}\] \[\Rightarrow \] \[{{h}^{2}}+{{k}^{2}}-\frac{2h}{3}({{x}_{1}}+{{x}_{2}}+{{x}_{3}})\] \[-\frac{2k}{3}({{y}_{1}}+{{y}_{2}}+{{y}_{3}})+\sum\limits_{i=1}^{3}{(x_{i}^{2}+y_{i}^{2})}-c=0\] So, locus of\[(h,k)\]is \[{{x}^{2}}+{{y}^{2}}-\frac{2x}{3}({{x}_{1}}+{{x}_{2}}+{{x}_{3}})\] \[-\frac{2y}{3}({{y}_{1}}+{{y}_{2}}+{{y}_{3}})+\lambda =0\] where \[\lambda =\sum\limits_{i=1}^{3}{(x_{i}^{2}+y_{i}^{2})-c=0}\]constant Clearly the locus of the circle with, centre at \[\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)\].
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