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MHT CET202520 Apr 2025Morning ShiftMathematicsCircleActual

The equation of the circle passing through the point (1,1) and having two diameters along the pair of lines x^2-y^2-2 x+4 y-3=0 is

Options

  1. A(x+2)^2+(y-2)^2=4
  2. B(x-3)^2+(y-1)^2=4
  3. C(x-1)^2+(y-2)^2=1
  4. D(x+1)^2+(y+2)^2=1

Correct answer

C. (x-1)^2+(y-2)^2=1

Step-by-step solution

The equation x^2 - y^2 - 2x + 4y - 3 = 0 represents a pair of intersecting lines, which are diameters of the circle. Since B^2 - 4AC = 0 - 4(1)(-1) = 4 > 0 and the condition for a pair of straight lines is satisfied, we factor it into two linear equations. Assume the lines take the form (x - y + c₁)(x + y + c₂) = 0 . Expanding gives x^2 - y^2 + (c₁ + c₂)x + (c₁ - c₂)y + c₁c₂ = 0 . Matching coefficients with the original equation yields: c₁ + c₂ = -2 c₁ - c₂ = 4 c₁c₂ = -3 Adding the first two equations gives 2c₁ = 2

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