Show that the points (5,5),(6,4),(-2,4) and(7,1) are concyclic. Find the equation, radius and centre of this circle.

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$$\begin{gathered} \mathrm{If} \mathrm{four} \mathrm{points} \mathrm{are} \mathrm{concyclic}, \mathrm{then} 4\mathrm{th} \mathrm{point} \mathrm{must} \mathrm{satisfy} \mathrm{equation} \mathrm{of} \mathrm{circle} \\ \mathrm{passing} \mathrm{through} \mathrm{other} \mathrm{three} \mathrm{points}. \\ \mathrm{Let} {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0 \mathrm{be} \mathrm{equation} \mathrm{of} \mathrm{circle}. \\ \left(5,5\right), \left(6,4\right) \mathrm{and} \left(7,1\right) \mathrm{must} \mathrm{satisfy} \mathrm{circle} \\ \left(5,5\right)\to 25+25+10\mathrm{g}+10\mathrm{f}+\mathrm{c}=0 \\ \Rightarrow 10\mathrm{g}+10\mathrm{f}+\mathrm{c}=-50...\left(\mathrm{i}\right) \\ \left(6,4\right)\to 36+16+12\mathrm{g}+8\mathrm{f}+\mathrm{c}=0 \\ \Rightarrow 12\mathrm{g}+8\mathrm{f}+\mathrm{c}=-52...\left(\mathrm{ii}\right) \\ \left(7,1\right)\to 49+1+14\mathrm{g}+2\mathrm{f}+\mathrm{c}=0 \\ \Rightarrow 14\mathrm{g}+2\mathrm{f}+\mathrm{c}=-50...\left(\mathrm{iii}\right) \\ \left(\mathrm{ii}\right)-\left(\mathrm{i}\right) \\ 2\mathrm{g}-2\mathrm{f}=-2...\left(\mathrm{iv}\right) \\ \left(\mathrm{iii}\right)-\left(\mathrm{ii}\right) \\ \Rightarrow 2\mathrm{g}-6\mathrm{f}=2..\left(\mathrm{v}\right) \\ \left(\mathrm{iv}\right)-\left(\mathrm{v}\right) \\ 2\mathrm{g}-2\mathrm{f}-2\mathrm{g}+6\mathrm{f}=-2-2 \\ \Rightarrow 4\mathrm{f}=-4 \\ \Rightarrow \mathrm{f}=-1 \\ \mathrm{Put} \mathrm{in} \left(\mathrm{iv}\right) \\ 2\mathrm{g}+2=-2 \\ \Rightarrow \mathrm{g}=-2 \\ 10\mathrm{g}+10\mathrm{f}+\mathrm{c}=-50 \\ \mathrm{Put} \mathrm{in} \left(\mathrm{i}\right) \\ -20-10+\mathrm{c}=-50 \\ \Rightarrow \mathrm{c}=-20 \\ \mathrm{Hence} \mathrm{equation} \mathrm{is} \\ {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-2\mathrm{y}-20=0 \\ \mathrm{Put} \left(-2,4\right) \mathrm{in} \mathrm{R}.\mathrm{H}.\mathrm{S}. \\ 4+16+8-8-20=28-28=0 \\ \mathrm{Hence} \left(-2,4\right) \mathrm{satisfies} \mathrm{circle} \mathrm{passing} \mathrm{through} \left(5,5\right), \left(6,4\right) \mathrm{and} \left(7,1\right) \\ \mathrm{Hence} \\ \left(5,5\right), \left(6,4\right),\left(-2,4\right) \mathrm{and} \left(7,1\right) \mathrm{are} \mathrm{concyclic}. \\ \mathrm{Centre} \mathrm{is} \left(2,1\right) \\ \mathrm{Radius}=\sqrt{4+1-\left(-20\right)}=\sqrt{25}=5 \end{gathered}$$

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