Question

The shortest distance from the line 3x + 4y = 25 to the circle x²+y²=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

The equation of the circle is $x^2 + y^2 = 6x - 8y$. Rearranging this to the standard form $(x-h)^2 + (y-k)^2 = r^2$, we get $(x-3)^2 + (y+4)^2 = 25$. Thus, the center of the circle is $C=(3, -4)$ and the radius is $r=5$.

The equation of the line is $3x + 4y = 25$, which can be written as $3x + 4y - 25 = 0$.

The perpendicular distance $d$ from the center of the circle $(3, -4)$ to the line $3x + 4y - 25 = 0$ is calculated using the formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$: $d = \frac{|3(3) + 4(-4) - 25|}{\sqrt{3^2 + 4^2}} = \frac{|9 - 16 - 25|}{\sqrt{9 + 16}} = \frac{|-32|}{5} = \frac{32}{5}$

Since the distance from the center to the line ($d = \frac{32}{5} = 6.4$) is greater than the radius ($r=5$), 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 $= d - r = \frac{32}{5} - 5 = \frac{32 - 25}{5} = \frac{7}{5}$.