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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 square units, the equation of the circle is

Options

  1. Ax^2+y^2+2 x-2 y-47=0
  2. Bx^2+y^2+2 x-2 y-62=0
  3. Cx^2+y^2-2 x+2 y-62=0
  4. Dx^2+y^2-2 x+2 y-47=0

Correct answer

D. x^2+y^2-2 x+2 y-47=0

Step-by-step solution

Point of intersection of 3 x-4 y-7=0 and 2 x-3 y-5=0 is (1,-1) , which is the centre of the circle and radius =7 . Equation is (x-1)^2+(y+1)^2=49 x^2+y^2-2 x+2 y-47=0

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