How to find the shortest distance from a line to circle while their equations are given Consider a line $L$ of equation $ 3x + 4y - 25 = 0 $ and a real circle $C$ of real center of equation $ x^2 + y^2 -6x +8y =0 $ 
I need to find the shortest distance from the line $L$ to the circle $C$.
How do I find that?
I am new to coordinate geometry of circles and line.
And I noted the slope of $L$ to be $\frac{-A}{B} = \frac{-3}{4} $ which means the line is inclined to $ -37° $ with $+x axis$ And circle centered at $ (3,-4) $ and of radius $5$ units.
By diagram, its difficult to figure out. Can we figure out easily by diagram or there is an algebraic way which is good for this? 
 A: Note that the shortest distance between a line and a circle will be the perpendicular distance of the line from the centre of the circle, minus the radius.
The circle can be written as $C=(x-3)^2+(y+4)^2=25=5^2$. So, $?$
A: Hint:
Any point on the circle can be set as $$P(3+5\cos t,5\sin t-4)$$
The distance of this point from $L$  will be $$\dfrac{|3(3+5\cos t)+4(5\sin t-4)-25|}{\sqrt{3^2+4^2}}$$
$3(3+5\cos t)+4(5\sin t-4)-25=5(3\cos t+4\sin t)-32$
Now $-\sqrt{3^2+4^2}\le3\cos t+4\sin t\le\sqrt{3^2+4^2}$
$\iff-5\cdot5-32\le5(3\cos t+4\sin t)-32\le5\cdot5-32$
$\iff7\le|3(3+5\cos t)+4(5\sin t-4)-25|\le57$
A: The shortest distance will be a perpendicular to the line which starts at your circle center $(3,-4)$ and is perpendicular to the line, so of slope $4/3.$ 
To continue, the perpendicular parametrically is $x=3+(4/3)t,\ y=-4+(4/3)t.$ Now plug that into the line equation, get $t$, and put that $t$ into the above parametric of the perpendicular to get the other end of the perpendicular.
Edit: In the parametrization of the perpendicular it should be $x=3+t$ not what I had, which would make the perpendicular have incorrect slope $1.$
