Question bank/JEE Main/Mathematics

Equation of the circle which circumscribes the square formed by
the lines xy - 9x - 2y + 18 = 0 and xy - 5x - 6y + 30 = 0 is

Ax^2 + y^2 - 4x - 10y + 21 = 0Bx^2 + y^2 - 8x - 14y + 57 = 0Cx^2 + y^2 - 12x - 10y + 53 = 0Dx^2 + y^2 - 7x - 15y + 64 = 0

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Given equations represent the lines x = 2, y = 9, x = 6, y = 5
which forms a square of each side equal to 4, vertex A(2, 5), C(6, 9).
Centre of the circle is the midpoint of AC and the radius is (1/2) AC.
Hence the equation of the circle is
$\Rightarrow$ x^2 + y^2 - 8x - 14y + 16 + 49 - 8 = 0
$\Rightarrow$ x^2 + y^2 - 8x - 14y + 57 =

Source: Grass-G / M-23 Mathematics Paper 31 Dec FST.docx