Given, circle is inscribed in square formed by the lines
$x^2$ - 8x + 12 = 0 and $y^2$ - 14y + 45 = 0
$\Rightarrow$ x = 6 and x = 2, y = 5 and y = 9
which could be plotted as
where, ABCD clearly forms a square.
$\therefore$ Centre of inscribed circle
= Point of intersection of diagonals
= Mid-point of AC or BD
$\Big( \frac{2+6}{2}\Big), \Big( \frac{5+9}{2}\Big)=(4,7)$
$\Rightarrow$ Centre of inscribed circle is (4, 7)