Find the equation of the circle passing through the point (7,3) having radius 3 units and whose center lie on a line y=x-1

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$$\begin{gathered} \mathrm{Let} \left(\mathrm{h},\mathrm{k}\right) \mathrm{be} \mathrm{the} \mathrm{coordinates} \mathrm{of} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{the} \mathrm{circle}. \\ \mathrm{Since} \mathrm{centre} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{lies} \mathrm{on} \mathrm{line} \mathrm{y}=\mathrm{x}-1, \mathrm{then} \mathrm{it} \mathrm{must} \mathrm{satisfy} \mathrm{it}. \\ \mathrm{Now}, \mathrm{k} = \mathrm{h} - 1 ....\left(1\right) \\ \mathrm{Now}, \mathrm{the} \mathrm{circle} \mathrm{is} \mathrm{passing} \mathrm{through} \mathrm{the} \mathrm{point} \left(7,3\right). \\ \mathrm{radius}=3 \mathrm{units} \\ \mathrm{distance} \mathrm{between} \left(\mathrm{h},\mathrm{k}\right) \mathrm{and} \left(7,3\right) = \mathrm{radius} \\ \Rightarrow \sqrt{{\left(\mathrm{h}-7\right)}^2+{\left(\mathrm{h}-1-3\right)}^2}=3 \\ \Rightarrow {\left(\mathrm{h}-7\right)}^2+{\left(\mathrm{h}-4\right)}^2=9 \\ \Rightarrow {\mathrm{h}}^2+49-14\mathrm{h}+{\mathrm{h}}^2+16-8\mathrm{h}=9 \\ \Rightarrow 2{\mathrm{h}}^2-22\mathrm{h}+65-9=0 \\ \Rightarrow 2{\mathrm{h}}^2-22\mathrm{h}+56=0 \\ \Rightarrow {\mathrm{h}}^2-11\mathrm{h}+28=0 \\ \Rightarrow {\mathrm{h}}^2-7\mathrm{h}-4\mathrm{h}+28=0 \\ \Rightarrow \mathrm{h}\left(\mathrm{h}-7\right)-4\left(\mathrm{h}-7\right)=0 \\ \Rightarrow \left(\mathrm{h}-7\right)\left(\mathrm{h}-4\right)=0 \\ \Rightarrow \mathrm{h}=7 \mathrm{or} \mathrm{h}=4 \\ \Rightarrow {\mathrm{h}}_1=7 \\ \Rightarrow {\mathrm{h}}_2=4 \\ \mathrm{hence} \mathrm{there} \mathrm{can} \mathrm{be} \mathrm{two} \mathrm{circles} \mathrm{having} \mathrm{the} \mathrm{coordinates} \mathrm{of} \mathrm{centre} \mathrm{as} : \\ {\mathrm{C}}_1=\left({\mathrm{h}}_1,{\mathrm{h}}_1-1\right)=\left(7,6\right) \\ {\mathrm{C}}_2=\left({\mathrm{h}}_2,{\mathrm{h}}_2-1\right)=\left(4,3\right) \\ \mathbf{hence}\mathbf{ }\mathbf{eqn}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{circles}\mathbf{ }\mathbf{are} \\ {\left(\mathbf{x}\mathbf{-}\mathbf{7}\right)}^{\mathbf{2}}\mathbf{+}{\left(\mathbf{y}\mathbf{-}\mathbf{6}\right)}^{\mathbf{2}}\mathbf{=}\mathbf{9} \\ \mathbf{and} \\ {\left(\mathbf{x}\mathbf{-}\mathbf{4}\right)}^{\mathbf{2}}\mathbf{+}{\left(\mathbf{y}\mathbf{-}\mathbf{3}\right)}^{\mathbf{2}}\mathbf{=}\mathbf{9} \end{gathered}$$

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