Centre of the circle circumscribed in a rectangle formed by the lines $x^2-8x+12=0$and $y^2-14y+40=0$ is 
 

$$\begin{gathered} {\mathrm{x}}^2-8\mathrm{x}+12=0 \\ (\mathrm{x}-6)(\mathrm{x}-2)=0 \\ \mathrm{x}=6, \mathrm{x}=2 \\ {\mathrm{y}}^2-14\mathrm{y}+40=0 \\ (\mathrm{y}-4)(\mathrm{y}-10)=0 \\ \mathrm{y}=4, \mathrm{y}=10 \\ \mathrm{A}(6, 4) \mathrm{B}(2, 10) \\ \mathrm{C}=\mathrm{mid}-\mathrm{point} \mathrm{of} \mathrm{AB} \\ \mathrm{C}=(4, 7) \end{gathered}$$