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(A) The line $L_1: \lambda x - y + 1 = 0$ intersects the axes at $(0, 1)$ and $(-1/\lambda, 0)$.The line $L_2: x - 2y + 3 = 0$ intersects the axes at

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$2$

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(A) The line $L_1: \lambda x - y + 1 = 0$ intersects the axes at $(0, 1)$ and $(-1/\lambda, 0)$.The line $L_2: x - 2y + 3 = 0$ intersects the axes at $(0, 3/2)$ and $(-3, 0)$.For a circle to pass through these four points,the points must be concyclic.Four points $(x_1, 0), (x_2, 0), (0, y_1), (0, y_2)$ are concyclic if $x_1 x_2 = y_1 y_2$.Here,the x-intercepts are $-1/\lambda$ and $-3$,and the y-intercepts are $1$ and $3/2$.Thus,$(-1/\lambda) 	imes (-3) = 1 	imes (3/2)$.$3/\lambda = 3/2$.Therefore,$\lambda = 2$.

If a circle passes through the points of intersection of the lines $\lambda x - y + 1 = 0$ and $x - 2y + 3 = 0$ with the coordinate axes,then the value of $\lambda$ is:

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