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?

The squared distance from $(2,1)$ to an arbitrary point $(x,y)=(x,4x/3+20/3)$ of the straight line is
$$(x-2)^2+\left(\frac43x+\frac{20}{3}-1\right)^2=\frac{25}{9}x^2+\frac{100}{9}x+\frac{35}{9}.$$
As the vertex of that parabola occurs at $x=-2$, the square of the minimal distance is $25$, hence the distance is $5$.  Now subtract $2$.
