Let the circle S = x^2 + y^2 + 2gx + 2fy + c | vedclass

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(B) The equation of the circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the circle touches both the positive $X$-axis and the positive $Y$-axis, its

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$96$

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(B) The equation of the circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the circle touches both the positive $X$-axis and the positive $Y$-axis, its center must be $(r, r)$ and its radius must be $r$, where $r > 0$. Thus, the equation becomes $(x - r)^2 + (y - r)^2 = r^2$, which simplifies to $x^2 + y^2 - 2rx - 2ry + r^2 = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$, we get $g = -r$ and $f = -r$. Since the point $(2, 4)$ lies on the circle, we substitute it into the equation: $(2 - r)^2 + (4 - r)^2 = r^2$ $4 - 4r + r^2 + 16 - 8r + r^2 = r^2$ $r^2 - 12r + 20 = 0$ $(r - 10)(r - 2) = 0$ So, the two possible radii are $r_1 = 10$ and $r_2 = 2$. The areas of the two circles are $A_1 = \pi(10)^2 = 100\pi$ and $A_2 = \pi(2)^2 = 4\pi$. The difference of their areas is $100\pi - 4\pi = 96\pi$.

Let the circle $S = x^2 + y^2 + 2gx + 2fy + c = 0$ touch the positive $X$-axis and the positive $Y$-axis. Let $(2, 4)$ be a point on the circle $S = 0$. If two such circles exist, then the difference of their areas is (in $\pi$)

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