The centre of circle inscribed in square formed by the line $x^2-8x+12=0$ and $y^2-14y+45=0$ is

$x^2-8x+12=0 \Rightarrow (x-6)(x-2)=0$

$y^2-14y+45=0 \Rightarrow (y-5)(y-9)=0$

The sides of a square are x = 2, x = 6, y = 5, y = 9

Then centre of circle inscribed in square will be

$(\frac{2+6}{2},\frac{5+9}{2})=(4,7)$