the centre of circle inscribed in the square formed by the lines x square-8x+12=0 and ysquare-14y+45=0

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\mathrm{x}}^2-8\mathrm{x}+12=0 \\ \Rightarrow {\mathrm{x}}^2-6\mathrm{x}-2\mathrm{x}+12=0 \\ \Rightarrow \mathrm{x}\left(\mathrm{x}-6\right)-2\left(\mathrm{x}-6\right)=0 \\ \Rightarrow \left(\mathrm{x}-2\right)\left(\mathrm{x}-6\right)=0 \\ \mathrm{x}=2 \mathrm{and} \mathrm{x}=6 \mathrm{are} \mathrm{two} \mathrm{lines} \\ {\mathrm{y}}^2-14\mathrm{y}+45=0 \\ \Rightarrow {\mathrm{y}}^2-5\mathrm{y}-9\mathrm{y}+45=0 \\ \Rightarrow \mathrm{y}\left(\mathrm{y}-5\right)-9\left(\mathrm{y}-5\right)=0 \\ \Rightarrow \left(\mathrm{y}-5\right)\left(\mathrm{y}-9\right)=0 \\ \mathrm{y}=5 \mathrm{and} \mathrm{y}=9 \mathrm{are} \mathrm{other} \mathrm{lines} \\ \mathrm{Let} \mathrm{us} \mathrm{draw} \mathrm{lines} \mathrm{on} \mathrm{graph}. \end{gathered}$$





$$\begin{gathered} \mathrm{Centre} \mathrm{of} \mathrm{square} \mathrm{will} \mathrm{be} \mathrm{centre} \mathrm{of} \mathrm{cirlce}. \\ \mathrm{Centre} \mathrm{of} \mathrm{square}=\left(\frac{2+6}{2},\frac{5+9}{2}\right)=\left(4,7\right) \left[\mathrm{Answer}\right] \end{gathered}$$

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