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

$$\begin{gathered} 2\mathrm{x}-\mathrm{y}+1=0 -\left(1\right) \\ \mathrm{x}+\mathrm{\lambda y}-3=0 -\left(2\right) \\ \mathrm{From} \mathrm{eq} -\left(1\right) \\ \mathrm{put} \mathrm{x}=0 then \mathrm{y}=1 \\ Let A=(0, 1) \\ put \mathrm{y}=0 then \mathrm{x}=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ Let B=\left(\frac{-1}{2}, 0\right) \end{gathered}$$

$$\begin{gathered} \mathrm{From} \mathrm{eq}-\left(2\right) \\ \mathrm{put} \mathrm{x}=0 then \mathrm{y}=\frac{3}{\mathrm{\lambda }} \\ Let C= \left(0, \frac{3}{\mathrm{\lambda }}\right) \\ and put \mathrm{y}=0 then \mathrm{x}=3 \\ Let D= (3, 0) \\ If A,B,C,D are concyclic points then \\ OA.OC=OB.OD \\ \Rightarrow 1\left(\frac{3}{\lambda }\right)=\left(\frac{1}{2}\right)3 \\ \Rightarrow \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$\mathrm{\lambda }$}\right.-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=0 \\ \Rightarrow \mathrm{\lambda }=2 \end{gathered}$$