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JEE MainMathematicsCircle

The equation ( - 1) x^2 + ( - 2) xy + 2 y^2 + 8x - 12y + = 0 represents a circle C₁ . It is given that C₁ intersects the circle C₂: x^2 + y^2 + 6x + 2y - 4 = 0 orthogonally. The value of + + is:

Options

  1. A15
  2. B37
  3. C25
  4. D1

Correct answer

C. 25

Step-by-step solution

For the general second-degree equation to represent a circle, the coefficient of xy must be zero and the coefficients of x^2 and y^2 must be equal. Coefficient of xy = 0 - 2 = 0 = 2 . Coefficient of x^2 = Coefficient of y^2 - 1 = 2 = 3 . Substituting these values, the equation of C₁ becomes: 2x^2 + 2y^2 + 8x - 12y + = 0 Dividing by 2 to convert it to standard form: x^2 + y^2 + 4x - 6y + 2 = 0 Here, g₁ = 2 , f₁ = -3 , and c₁ = 2 . For the circle C₂: x^2 + y^2 + 6x + 2y - 4 = 0 , we have: g₂ = 3 , f₂ = 1 , and c₂ = -

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