If the lines $2 x + 3 y + 1 = 0, 6 x + 4 y + 1 = 0$ intersect the co-ordinate 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

Two lines $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ intersect the axes at four points, the conditon that the four points are to be concyclic is $a_{1} b_{2} + a_{2} b_{1} = 0$

The equation of the circle is $(a_{1} x + b_{1} y + c_{1}) (a_{2} x + b_{2} y + c_{2}) - (a_{1} b_{2} + a_{2} b_{1}) x y = 0$

Eqaution of the required circle is

$\begin{array}{rcl} (2 x + 3 y + 1) (6 x + 4 y + 1) - x y (8 + 18) & = & 0 \\ 12 x^{2} + 12 y^{2} + 8 x + 7 y + 1 & = & 0 \end{array}$

Therefore, the equaiton of the circle passing through the concyclic points is $\begin{array}{rcl} 12 x^{2} + 12 y^{2} + 8 x + 7 y + 1 = 0 \end{array}$