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(D) The given curve is $|x - 1| + |y - 4| = 6$. This represents a square with its center at $(1, 4)$.The sides of the square are given by the lines $\

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$x^2 + y^2 - 2x - 8y - 1 = 0$

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(D) The given curve is $|x - 1| + |y - 4| = 6$. This represents a square with its center at $(1, 4)$.The sides of the square are given by the lines $\pm(x - 1) \pm(y - 4) = 6$. One such side is $(x - 1) + (y - 4) = 6$,which simplifies to $x + y = 11$,or $x + y - 11 = 0$.The radius $r$ of the circle touching this curve is the perpendicular distance from the center $(1, 4)$ to the line $x + y - 11 = 0$:$r = \left| \frac{1 + 4 - 11}{\sqrt{1^2 + 1^2}} \right| = \left| \frac{-6}{\sqrt{2}} \right| = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.The equation of the circle with center $(1, 4)$ and radius $r = 3\sqrt{2}$ is:$(x - 1)^2 + (y - 4)^2 = (3\sqrt{2})^2$$(x^2 - 2x + 1) + (y^2 - 8y + 16) = 18$$x^2 + y^2 - 2x - 8y + 17 = 18$$x^2 + y^2 - 2x - 8y - 1 = 0$.

The equation of a circle touching the curve $|x - 1| + |y - 4| = 6$ is

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