question_answer

Three distinct points A, B and C are given in the 2-dimensional coordinate plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point (-1, 0) is equal to\[\frac{1}{3}\]. Then the circumcentre of the triangle ABC is at the point     AIEEE  Solved  Paper-2009

A) (0, 0)                     

B) \[\left( \frac{5}{4},0 \right)\]     

C)        \[\left( \frac{5}{2},0 \right)\]     

D)         \[\left( \frac{5}{3},0 \right)\]

Correct Answer: B

Solution :

Let the point A is (h, k) \[\frac{AP}{AQ}=\frac{1}{3}\] \[3AP=AQ\] \[\Rightarrow \]\[9A{{P}^{2}}=A{{Q}^{2}}\] \[\Rightarrow \]\[9[{{(h1)}^{2}}+{{k}^{2}}]={{(h1)}^{2}}+{{k}^{2}}\] \[\Rightarrow \]\[{{h}^{2}}+{{k}^{2}}-\frac{5}{2}h-1=0\] Locus of\[A(h,\text{ }k)\]is \[{{x}^{2}}+{{y}^{2}}-\frac{5}{2}x-1=0\] \[\therefore \] Circumcentre of\[\Delta ABC\]is \[\left( \frac{5}{4},0 \right)\]