The pair of lines x^2 - 8x + 12 = 0 and y^2 - | vedclass

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(C) The given equations are $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$. Solving $x^2 - 8x + 12 = 0$: $(x - 6)(x - 2) = 0 \Rightarrow x = 2, x = 6$.

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$(4, 7)$

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(C) The given equations are $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$. Solving $x^2 - 8x + 12 = 0$: $(x - 6)(x - 2) = 0 \Rightarrow x = 2, x = 6$. Solving $y^2 - 14y + 45 = 0$: $(y - 9)(y - 5) = 0 \Rightarrow y = 5, y = 9$. The vertices of the square are $A(2, 5)$,$B(2, 9)$,$C(6, 9)$,and $D(6, 5)$. The center of the inscribed circle is the midpoint of the diagonals of the square. Midpoint of diagonal $AC = \left(\frac{2 + 6}{2}, \frac{5 + 9}{2}\right) = \left(\frac{8}{2}, \frac{14}{2}\right) = (4, 7)$.

The pair of lines $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$ form a square. What is the center of the circle inscribed in the square?

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