The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq units. Then the equation of the circle is: - Mathematics MCQ AIEEE-CBSE-ENG-03 2026 | ExamTest.live

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The lines 2 x - 3 y = 5 and 3 x - 4 y = 7 are diameters of a circle having area as 154 sq units. Then the equation of the circle is:

Solution & Step-by-step Explanation

The intersection of diameters is the center.Solving 2 x - 3 y = 5 and 3 x - 4 y = 7 :Multiply first by 3 and second by 2: 6 x - 9 y = 15 6 x - 8 y = 14 Subtracting: - y = 1 ⟹ y = - 1 . 2 x - 3 (- 1) = 5 ⟹ 2 x = 2 ⟹ x = 1 .Center is (1, - 1) .Area = π R^{2} = 154 . \frac{22}{7} R^{2} = 154 ⟹ R^{2} = 154 \cdot \frac{7}{22} = 7 \cdot 7 = 49 .Equation: (x - 1)^{2} + (y + 1)^{2} = 49 $ x^{2} - 2 x + 1 + y^{2} + 2 y + 1 = 49 ⟹ x^{2} + y^{2} - 2 x + 2 y = 47 $

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The lines 2 x - 3 y = 5 and 3 x - 4 y = 7 are diameters of a circle having area as 154 sq units. Then the equation of the circle is:

x^{2} + y^{2} + 2 x - 2 y = 62

x^{2} + y^{2} + 2 x - 2 y = 47

x^{2} + y^{2} - 2 x + 2 y = 47

x^{2} + y^{2} - 2 x + 2 y = 62

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