The vertices A,B,C of a triangle ABC have coordinates (4,4),(5,3),(6,0) respectively. Find the equations of the perpendicular bisectors of AB and BC ,the coordinates of the circumvented and the equation of the circumcircle of triangle ABC

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \mathrm{A}\left(4,4\right), \mathrm{B}\left(5,3\right) \mathrm{and} \mathrm{C}\left(6,0\right) \\ \mathrm{Slope} \mathrm{of} \mathrm{AB}=\frac{3-4}{5-4}=-1 \\ \Rightarrow \mathrm{Slope} \mathrm{of} \mathrm{perpendicular} \mathrm{bisector} \mathrm{of} \mathrm{AB}={\mathrm{m}}_1=-\frac{1}{\mathrm{Slope} \mathrm{of} \mathrm{AB}}=-\frac{1}{\left(-1\right)}=1 \\ \mathrm{Perpendicular} \mathrm{bisector} \mathrm{of} \mathrm{AB} \mathrm{will} \mathrm{pass} \mathrm{through} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{AB}. \\ \mathrm{Mid} \mathrm{point} \mathrm{of} \mathrm{AB} \left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left(\frac{5+4}{2},\frac{4+3}{2}\right)=\left(\frac{9}{2},\frac{7}{2}\right) \\ \mathrm{Hence} \mathrm{equation} \mathrm{of} \mathrm{perpendicular} \mathrm{bisector} \mathrm{of} \mathrm{AB} \mathrm{will} \mathrm{be}: \\ \mathrm{y}-{\mathrm{y}}_1={\mathrm{m}}_1\left(\mathrm{x}-{\mathrm{x}}_1\right) \\ \Rightarrow \mathrm{y}-\frac{7}{2}=\left(\mathrm{x}-\frac{9}{2}\right) \\ \Rightarrow \frac{2\mathrm{y}-7}{2}=\frac{2\mathrm{x}-9}{2} \\ \Rightarrow 2\mathrm{y}-7=2\mathrm{x}-9 \\ \Rightarrow 2\mathrm{x}-2\mathrm{y}=2 \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{1}\mathbf{ };\left(\mathrm{i}\right) \\ \mathrm{Slope} \mathrm{of} \mathrm{BC}=\frac{0-3}{6-5}=-3 \\ \Rightarrow \mathrm{Slope} \mathrm{of} \mathrm{perpendicular} \mathrm{bisector} \mathrm{of} \mathrm{BC}={\mathrm{m}}_2=-\frac{1}{\left(-3\right)}=\frac{1}{3} \\ \mathrm{Perpendicular} \mathrm{bisector} \mathrm{of} \mathrm{BC} \mathrm{will} \mathrm{pass} \mathrm{through} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{AB}. \\ \mathrm{Mid} \mathrm{point} \mathrm{ofBC} \left({\mathrm{x}}_2,{\mathrm{y}}_2\right)=\left(\frac{5+6}{2},\frac{3+0}{2}\right)=\left(\frac{11}{2},\frac{3}{2}\right) \\ \mathrm{Hence} \mathrm{equation} \mathrm{of} \mathrm{perpendicular} \mathrm{bisector} \mathrm{of} \mathrm{BC} \mathrm{will} \mathrm{be}: \\ \mathrm{y}-{\mathrm{y}}_2={\mathrm{m}}_2\left(\mathrm{x}-{\mathrm{x}}_2\right) \\ \Rightarrow \mathrm{y}-\frac{3}{2}=\frac{1}{3}\left(\mathrm{x}-\frac{11}{2}\right) \\ \Rightarrow \frac{2\mathrm{y}-3}{2}=\frac{2\mathrm{x}-11}{6} \\ \Rightarrow 2\mathrm{y}-3=\frac{2\mathrm{x}-11}{3} \\ \Rightarrow 6\mathrm{y}-9=2\mathrm{x}-11 \\ \Rightarrow 2\mathrm{x}-6\mathrm{y}=2 \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{1} ;\left(\mathrm{ii}\right) \\ \mathrm{Point} \mathrm{of} \mathrm{intersection} \mathrm{of} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \mathrm{will} \mathrm{give} \mathrm{circumcentre}. \\ \left(\mathrm{i}\right)-\left(\mathrm{ii}\right) \\ \mathrm{x}-\mathrm{y}-\mathrm{x}+3\mathrm{y}=1-1 \\ \Rightarrow \mathrm{y}=0 \\ \mathrm{Put} \mathrm{in} \left(\mathrm{i}\right) \\ \mathrm{x}=1 \\ \mathbf{Hence}\mathbf{ }\mathbf{circumcentre}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{O}\left(\mathbf{1}\mathbf{,}\mathbf{0}\right) \\ \mathrm{Radius} \mathrm{of} \mathrm{circumcentre}=\mathrm{OA}=\mathrm{OB}=\mathrm{OC} \\ \mathrm{C}\left(6,0\right) \\ \mathrm{r}=\mathrm{OC}=\sqrt{{\left(6-1\right)}^2+{\left(0-0\right)}^2}=5 \mathrm{units} \\ \mathrm{Hence} \mathrm{equation} \mathrm{of} \mathrm{circumcircle} \mathrm{will} \mathrm{be}: \\ {\left(\mathrm{x}-1\right)}^2+{\mathrm{y}}^2=5^2 \\ \mathbf{\Rightarrow }{\left(\mathbf{x}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{25} \end{gathered}$$
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