$\begin{array}{1 1}(A)\;(4,7) \\(B)\;(7,4)\\(C)\;(9,4) \\(D)\;(4,9) \end{array}$

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$x^2-8x+12=0$

=> $(x-6)(x-2)=0$

$y^2-14y+45=0$

=> $(y-5)(y-9)=0$

Thus sides of square are

$x=2,x=6,y=5,y=9$

Then centre of circle inscribed in square will be $ \bigg( \large\frac{2+6}{2}, \frac{5+9}{2}\bigg)$

=> $ \bigg( \large\frac{8}{2},\frac{14}{2}\bigg)$

=> $(4,7)$

Hence A is the correct answer.

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