Three distinct points A, B and C are given such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point - Mathematics MCQ AIEEE 2009 2026 | ExamTest.live

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Three distinct points A, B and C are given 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 1/3 . Then the circumcentre of the triangle A BC is at the point:

Solution & Step-by-step Explanation

Let a point be (x, y) . The condition is \frac{( x - 1 ) ^{2} + y ^{2}}{( x + 1 ) ^{2} + y ^{2}} = 1/3 . 9 [(x - 1)^{2} + y^{2}] = (x + 1)^{2} + y^{2} 9 [x^{2} - 2 x + 1 + y^{2}] = x^{2} + 2 x + 1 + y^{2} 8 x^{2} - 18 x - 2 x + 8 y^{2} + 8 = 0 8 x^{2} + 8 y^{2} - 20 x + 8 = 0 ⟹ x^{2} + y^{2} - \frac{5}{2} x + 1 = 0 .This represents a circle. Points A, B, C lie on this circle.The circumcentre of △ A BC is the centre of this circle.Centre = (- g, - f) = (5/4, 0) .

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Three distinct points A, B and C are given 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 1/3 . Then the circumcentre of the triangle A BC is at the point:

(0, 0)

(5/4, 0)

(2/5, 0)

(5/3, 0)

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