If the lines 3x - 4y - 7 = 0 and 2x - 3y - 5 = 0 are two diameters of a circle of area 49π square units, the equation of the circle is: - Mathematics MCQ AIEEE 2006 2026 | ExamTest.live

mediumMCQAIEEE 20062026Mathematics

1 attempts0% success rate1 mark

If the lines 3 x - 4 y - 7 = 0 and 2 x - 3 y - 5 = 0 are two diameters of a circle of area 49 π square units, the equation of the circle is:

Solution & Step-by-step Explanation

The intersection of two diameters is the center of the circle.Solving 3 x - 4 y = 7 and 2 x - 3 y = 5 :Multiply first by 2 and second by 3 : 6 x - 8 y = 14 6 x - 9 y = 15 Subtracting gives y = - 1 .Substitute y = - 1 into 2 x - 3 (- 1) = 5 ⟹ 2 x = 2 ⟹ x = 1 .Center is (1, - 1) .Area = π r^{2} = 49 π ⟹ r^{2} = 49 ⟹ r = 7 .Equation: (x - 1)^{2} + (y + 1)^{2} = 7^{2} x^{2} - 2 x + 1 + y^{2} + 2 y + 1 = 49 x^{2} + y^{2} - 2 x + 2 y - 47 = 0 .

Try it yourself before checking the explanation above.

If the lines 3 x - 4 y - 7 = 0 and 2 x - 3 y - 5 = 0 are two diameters of a circle of area 49 π square units, the equation of the circle is:

x^{2} + y^{2} + 2 x - 2 y - 47 = 0

x^{2} + y^{2} + 2 x - 2 y - 62 = 0

x^{2} + y^{2} - 2 x + 2 y - 62 = 0

x^{2} + y^{2} - 2 x + 2 y - 47 = 0

Practice with timed mock tests and track your performance across Mathematics.