The centre of the circle inscribed in the squ | vedclass

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Question

(A) The given equations are $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$.Solving $x^2 - 8x + 12 = 0$,we get $(x - 2)(x - 6) = 0$,so $x = 2$ and $x = 6

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$(4, 7)$

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(A) The given equations are $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$.Solving $x^2 - 8x + 12 = 0$,we get $(x - 2)(x - 6) = 0$,so $x = 2$ and $x = 6$.Solving $y^2 - 14y + 45 = 0$,we get $(y - 5)(y - 9) = 0$,so $y = 5$ and $y = 9$.The lines forming the square are $x = 2, x = 6, y = 5,$ and $y = 9$.The centre of the inscribed circle is the midpoint of the square,which is the intersection of the diagonals.The centre is given by $(\frac{2 + 6}{2}, \frac{5 + 9}{2}) = (\frac{8}{2}, \frac{14}{2}) = (4, 7)$.

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

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