The shortest distance from the line $3 x+4 y=25$ to the circle $x^2+y^2-6 x+8 y=0$ is
The shortest distance from the line $3 x+4 y=25$ to the circle $x^2+y^2-6 x+8 y=0$ is
$\frac{9}{5}$
$\frac{7}{5}$
$\frac{8}{5}$
$\frac{13}{5}$
Solution
We have a line $3 x+4 y=25$ and a circle $x^2+y^2-6 x+8 y=0$ for shortest distance. We draw a perpendicular from $\mathrm{C}=$ $(3,-4)$ to the straight line.
$\Rightarrow \mathrm{d}=\frac{|9-16-25|}{\sqrt{9+16}}=\frac{32}{5}$
hence shortcut distance $=\frac{32}{5}-5=\frac{7}{5}$