Question:

**Step 1: Convert the circle equation to standard form.** The given equation for the circle is $x^2 + y^2 = 6x - 8y$. Rearranging the terms to group $x$ and $y$: $x^2 - 6x + y^2 + 8y = 0$ To convert this to standard form, complete the square for the $x$ terms and $y$ terms: $(x^2 - 6x + 9) + (y^2 + 8y + 16) = 9 + 16$ This simplifies to: $(x - 3)^2 + (y + 4)^2 = 25$ From the standard form $(x - h)^2 + (y - k)^2 = r^2$, the center of the circle is $C = (3, -4)$ and the radius is $r = \sqrt{25} = 5$. **Step 2: Find the perpendicular distance from the center to the line.** The equation of the line is $3x + 4y = 25$, which can be written as $3x + 4y - 25 = 0$. The distance $d$ from a point $(x_0, y_0)$ to a line $ax + by + c = 0$ is given by the formula: $$d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$$ Using the center of the circle $C = (3, -4)$ as $(x_0, y_0)$ and the line $3x + 4y - 25 = 0$ (where $a=3$, $b=4$, $c=-25$): $$d = \frac{|3(3) + 4(-4) - 25|}{\sqrt{3^2 + 4^2}}$$ $$d = \frac{|9 - 16 - 25|}{\sqrt{9 + 16}}$$ $$d = \frac{|-32|}{\sqrt{25}}$$ $$d = \frac{32}{5}$$ **Step 3: Find the shortest distance from the line to the circle.** The distance from the center of the circle to the line is $d = \frac{32}{5}$. The radius of the circle is $r = 5$. Since $d = \frac{32}{5} = 6.4$ and $r = 5$, we have $d > r$. This indicates that the line does not intersect the circle; it lies entirely outside the circle. The shortest distance from an external line to a circle is the distance from the center to the line minus the radius of the circle. Shortest distance $= d - r$ Shortest distance $= \frac{32}{5} - 5$ Shortest distance $= \frac{32}{5} - \frac{25}{5}$ Shortest distance $= \frac{7}{5}$