Question

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

• A. 7/5
• B. 9/5
• C. 11/5
• D. 32/5

Answer 1

Answer: (A)

7/5

Answer 2

1. Standardize the Circle Equation: The given equation of the circle is $x^2 + y^2 = 6x - 8y$. Rearrange and complete the square to find the center $(h, k)$ and radius $r$: $x^2 - 6x + y^2 + 8y = 0$ $(x^2 - 6x + 9) + (y^2 + 8y + 16) = 9 + 16$ $(x-3)^2 + (y+4)^2 = 25$ Thus, the center of the circle is $C(3, -4)$ and the radius is $r = \sqrt{25} = 5$.

2. Standardize the Line Equation: The equation of the line is $3x + 4y = 25$. Rewrite it in the general form $Ax + By + C = 0$: $3x + 4y - 25 = 0$

3. Calculate Distance from Center to Line: The shortest distance 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 $C(3, -4)$ and the line $3x + 4y - 25 = 0$ ($A=3, B=4, C=-25, x_0=3, y_0=-4$): $d = \frac{|3(3) + 4(-4) - 25|}{\sqrt{3^2 + 4^2}}$ $d = \frac{|9 - 16 - 25|}{\sqrt{9 + 16}}$ $d = \frac{|-32|}{\sqrt{25}}$ $d = \frac{32}{5}$

4. Determine Shortest Distance to 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 $d > r$ ($\frac{32}{5} = 6.4 > 5$), the line does not intersect the circle. The shortest distance from the line to the circle is the distance from the center to the line minus the radius. Shortest Distance $= d - r$ Shortest Distance $= \frac{32}{5} - 5$ Shortest Distance $= \frac{32}{5} - \frac{25}{5}$ Shortest Distance $= \frac{7}{5}$