Find the equation of the circle passing through the vertices of the triangle whose sides are x+y-4=0 , x-y=2 , and 2x-y-2=0

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ {\mathrm{L}}_1:\mathrm{x}+\mathrm{y}=4 ;\left(\mathrm{i}\right) \\ {\mathrm{L}}_2:\mathrm{x}-\mathrm{y}=2 ;\left(\mathrm{ii}\right) \\ {\mathrm{L}}_3:2\mathrm{x}-\mathrm{y}=2 ;\left(\mathrm{iii}\right) \\ \left(\mathrm{i}\right)+\left(\mathrm{ii}\right) \\ \mathrm{x}+\mathrm{y}+\mathrm{x}-\mathrm{y}=6 \\ \Rightarrow \mathrm{x}=3 \\ \mathrm{Put} \mathrm{this} \mathrm{in} \left(\mathrm{i}\right) \\ 3+\mathrm{y}=4 \\ \Rightarrow \mathrm{y}=1 \\ \Rightarrow \mathrm{A}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left(3,1\right) \\ \left(\mathrm{iii}\right)-\left(\mathrm{ii}\right) \\ 2\mathrm{x}-\mathrm{y}-\mathrm{x}+\mathrm{y}=2-2 \\ \Rightarrow \mathrm{x}=0 \\ \mathrm{Put} \mathrm{this} \mathrm{in} \left(\mathrm{ii}\right) \\ 0-\mathrm{y}=2 \\ \mathrm{or} \mathrm{y}=-2 \\ \Rightarrow \mathrm{B}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)=\left(0,-2\right) \\ \left(\mathrm{iii}\right)+\left(\mathrm{i}\right) \\ 2\mathrm{x}-\mathrm{y}+\mathrm{x}+\mathrm{y}=6 \\ \mathrm{x}=2 \\ \mathrm{Put} \mathrm{x}=2 \mathrm{in} \left(\mathrm{i}\right) \\ 2+\mathrm{y}=4 \\ \Rightarrow \mathrm{y}=2 \\ \mathrm{C}\left({\mathrm{x}}_3,{\mathrm{y}}_3\right)=\left(2,2\right) \end{gathered}$$



$$\begin{gathered} \mathrm{Slope} \mathrm{of} {\mathrm{L}}_1\left({\mathrm{m}}_1\right)=-\frac{\mathrm{Coefficient} \mathrm{of} \mathrm{x}}{\mathrm{Coefficient} \mathrm{of} \mathrm{y}}=-\frac{1}{1}=-1 \\ \mathrm{Slope} \mathrm{of} {\mathrm{L}}_2\left({\mathrm{m}}_2\right)=-\frac{\mathrm{Coefficient} \mathrm{of} \mathrm{x}}{\mathrm{Coefficient} \mathrm{of} \mathrm{y}}=-\frac{1}{\left(-11\right)}=1 \\ \Rightarrow {\mathrm{m}}_1{\mathrm{m}}_2=-1 \\ \mathrm{Hence} {\mathrm{L}}_1 \mathrm{and} {\mathrm{L}}_2 \mathrm{are} \mathrm{perpendicular}. \\ \mathrm{As} \mathrm{angle} \mathrm{in} \mathrm{semi}-\mathrm{cirlce} \mathrm{is} 90°, \mathrm{hence} \mathrm{BC} \mathrm{will} \mathrm{be} \mathrm{the} \mathrm{diameter} \mathrm{of} \mathrm{the} \mathrm{circle}. \\ \mathrm{Hence} \mathrm{equation} \mathrm{of} \mathrm{circle} \mathrm{is} \mathrm{given} \mathrm{by}: \\ \left(\mathrm{x}-{\mathrm{x}}_2\right)\left(\mathrm{x}-{\mathrm{x}}_3\right)+\left(\mathrm{y}-{\mathrm{y}}_2\right)\left(\mathrm{y}-{\mathrm{y}}_3\right)=0 \\ \Rightarrow \left(\mathrm{x}-0\right)\left(\mathrm{x}-2\right)+\left(\mathrm{y}+2\right)\left(\mathrm{y}-2\right)=0 \\ \Rightarrow {\mathrm{x}}^2-2\mathrm{x}+{\mathrm{y}}^2-2\mathrm{y}+2\mathrm{y}-4=0 \\ \Rightarrow {\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{=}\mathbf{0} \end{gathered}$$
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