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(B) The points of intersection of the line $\lambda x - y + 1 = 0$ with the coordinate axes are $(-\frac{1}{\lambda}, 0)$ and $(0, 1)$.The points of i

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$2$

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(B) The points of intersection of the line $\lambda x - y + 1 = 0$ with the coordinate axes are $(-\frac{1}{\lambda}, 0)$ and $(0, 1)$.The points of intersection of the line $x - 2y + 3 = 0$ with the coordinate axes are $(-3, 0)$ and $(0, \frac{3}{2})$.Since the circle passes through these four points,it passes through $(0, 1), (-3, 0),$ and $(0, \frac{3}{2})$.The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Substituting $(0, 1)$,we get $1 + 2f + c = 0$.Substituting $(0, \frac{3}{2})$,we get $\frac{9}{4} + 3f + c = 0$.Subtracting these equations: $(3f - 2f) + (\frac{9}{4} - 1) = 0 \Rightarrow f = -\frac{5}{4}$.Then $c = -1 - 2(-\frac{5}{4}) = -1 + \frac{5}{2} = \frac{3}{2}$.Substituting $(-3, 0)$,we get $9 - 6g + \frac{3}{2} = 0$ $\Rightarrow 6g = \frac{21}{2}$ $\Rightarrow g = \frac{7}{4}$.The circle equation is $x^2 + y^2 + \frac{7}{2}x - \frac{5}{2}y + \frac{3}{2} = 0$.Since it passes through $(-\frac{1}{\lambda}, 0)$,we have $\frac{1}{\lambda^2} - \frac{7}{2\lambda} + \frac{3}{2} = 0$.Multiplying by $2\lambda^2$: $2 - 7\lambda + 3\lambda^2 = 0$.Solving $3\lambda^2 - 7\lambda + 2 = 0$,we get $(3\lambda - 1)(\lambda - 2) = 0$.Thus,$\lambda = 2$ or $\lambda = \frac{1}{3}$. Given the options,the correct value is $2$.

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

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