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If the lines $2 x + 3 y + 1 = 0, 6 x + 4 y + 1 = 0$ intersect the coordinate axes in 4 points, then the circle passing through the 4 points is

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$12 x^{2} + 12 y^{2} + 8 x + 7 y + 1 = 0$

$6 x^{2} + 6 y^{2} + 3 x + y = 0$

$12 x^{2} + 12 y^{2} + 8 x + 7 y + 3 = 0$

$x^{2} + y^{2} + 4 x - y + 3 = 0$

answer is A.

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Detailed Solution

$Reg circle is (2 x + 3 y + 1) (6 x + 4 y + 1) - x y term = 0$

$(2 x + 3 y + 1) (6 x + 4 y + 1) - x y term = 0$

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