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MHT CET202615 April 2026Morning ShiftMathematicsCircleActual

The centre and radius of the circle (a+1)x^2 + 3y^2 - 6x + 9y + a + 4 = 0 are respectively ...

Options

  1. A(-1, 3 2 ), 5 2
  2. B(-1, - 3 2 ), 5 2
  3. C(1, - 3 2 ), 5 2
  4. D(1, 3 2 ), 5 2

Correct answer

C. (1, - 3 2 ), 5 2

Step-by-step solution

For the given equation to represent a circle, the coefficients of x^2 and y^2 must be equal. a + 1 = 3 a = 2 Substituting a = 2 into the given equation: 3x^2 + 3y^2 - 6x + 9y + 6 = 0 Dividing by 3 , we get the standard form of the circle: x^2 + y^2 - 2x + 3y + 2 = 0 Comparing with x^2 + y^2 + 2gx + 2fy + c = 0 , we have 2g = -2 , 2f = 3 , and c = 2 . The centre of the circle is (-g, -f) = (1, - 3 2 ) . The radius of the circle is r = g^2 + f^2 - c = (-1)^2 + ( 3 2 )^2 - 2 = 1 + 9 4 - 2 = 5 4 = 5 2 . Answer: (1, - 3

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