The shortest distance from the line 3x + 4y = | vedclass

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(B) The equation of the circle is $x^2 + y^2 - 6x + 8y = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $C = (-g, -f) = (3,

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$\frac{7}{5}$

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(B) The equation of the circle is $x^2 + y^2 - 6x + 8y = 0$. Comparing this with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $C = (-g, -f) = (3, -4)$ and radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{3^2 + (-4)^2 - 0} = \sqrt{9 + 16} = 5$. The perpendicular distance $d$ from the center $(3, -4)$ to the line $3x + 4y - 25 = 0$ is given by $d = \frac{|3(3) + 4(-4) - 25|}{\sqrt{3^2 + 4^2}} = \frac{|9 - 16 - 25|}{\sqrt{25}} = \frac{|-32|}{5} = \frac{32}{5}$. The shortest distance from the line to the circle is $d - r = \frac{32}{5} - 5 = \frac{32 - 25}{5} = \frac{7}{5}$.

The shortest distance from the line $3x + 4y = 25$ to the circle $x^2 + y^2 - 6x + 8y = 0$ is

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