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?

Hint. You have shown that your line doesn't intersect the circle. Therefore the shortest distance between the circle and the line is given by the distance between that line and a line such that $$y=\frac43x+c,$$ where the last line is tangential to the circle -- in other words it intersects the circle in just one point. There are two such lines, and you can easily select the closest to the given line. Thus, you want to solve for values of $c$ such that $$(x-2)^2+\left(\frac43x+c-1\right)^2=4$$ has only one root. That is, set the discriminant of this quadratic in $x$ equal to $0.$ Then take the bigger value for $c.$ You then have your line. Compute the distance between this line and the given line, and you'd be done!
