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The equation of a circle passing through the origin is x^2+y^2-6 x+2 y=0 . The equation of one of its diameter is

Options

  1. Ax-3 y=0
  2. Bx+3 y=0
  3. Cx+y=0
  4. D3 x-y=0

Correct answer

B. x+3 y=0

Step-by-step solution

The given equation of the circle is x^2 + y^2 - 6x + 2y = 0 . Comparing this with the general equation of a circle x^2 + y^2 + 2gx + 2fy + c = 0 , we have 2g = -6 and 2f = 2 . Thus, g = -3 and f = 1 . The center of the circle is (-g, -f) = (3, -1) . A diameter of a circle is a line passing through its center. Therefore, the equation of the diameter must satisfy the coordinates of the center (3, -1) . Testing the options: For option (1): x - 3y = 3 - 3(-1) = 3 + 3 = 6 0 . For option (2): x + 3y = 3 + 3(-1) = 3 - 3 =

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