Shortest distance from circle to a line 
Let $C$ be a circle with center $(2, 1)$ and radius $2$. Find the shortest distance from the line $3y=4x+20$.

This should be very simple, but I seem to end up with no real solutions.
The shortest distance would be from the center of the circle perpendicular to the line right?
Solving the line for $y$ we get $y=\frac{4}{3}x+\frac{20}{3}$
Substituting this to the equation of the circle we get $(x-2)^2+(\frac{4}{3}x+\frac{20}{3}-1)^2=2^2$, but solving this for $x$ ended up with no real roots. What am I missing here?

What you would do is draw a perpendicular from the centre of the circle to that line.
That perpendicular passes through $C=(2,1)$ and is (obviously) perpendicular to $y=\dfrac{4}{3}x+\dfrac{20}{3}$
Line has a gradient of $-\dfrac{3}{4}$
$\begin{align} \dfrac{y-1}{x-2} &= -\dfrac{3}{4} \\ y&=-\dfrac{3}{4}x +\dfrac{5}{2}\end{align}$
The two lines intersect at $P=(-2, 4)$ which you can verify by solving this:
$\dfrac{4}{3}x+\dfrac{20}{3}=-\dfrac{3}{4}x +\dfrac{5}{2}$
The distance $PC$ is:
$\sqrt{(2-(-2))^2+(4-1)^2}=5$
So the shortest distance to the circumference is $PC$ minus the radius of the circle which is:
$5-2=\boxed{3}$
