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(A) The given equation of the circle is $x^2 + y^2 - 8x - 6y + 9 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g =

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$[ -\frac{7}{24}, \infty )$

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(A) The given equation of the circle is $x^2 + y^2 - 8x - 6y + 9 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -4$,$f = -3$,and $c = 9$.The center of the circle is $(4, 3)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{16 + 9 - 9} = \sqrt{16} = 4$.Let $m = \frac{y}{x}$,which implies $y = mx$ or $mx - y = 0$.For the line $mx - y = 0$ to intersect the circle,the perpendicular distance from the center $(4, 3)$ to the line must be less than or equal to the radius $r = 4$.Thus,$\frac{|m(4) - 3|}{\sqrt{m^2 + (-1)^2}} \le 4$.$|4m - 3| \le 4\sqrt{m^2 + 1}$.Squaring both sides,$(4m - 3)^2 \le 16(m^2 + 1)$.$16m^2 - 24m + 9 \le 16m^2 + 16$.$-24m \le 7$,which gives $m \ge -\frac{7}{24}$.Since the line $y = mx$ passes through the origin and the circle lies entirely in the first quadrant (as the center is $(4, 3)$ and radius is $4$,$x$ ranges from $0$ to $8$ and $y$ from $-1$ to $7$),the slope $m$ can take values from $-\frac{7}{24}$ to $\infty$.

If a variable point $(x, y)$ satisfies the equation $x^2 + y^2 - 8x - 6y + 9 = 0$,then the range of $\frac{y}{x}$ is

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