<s>[INST] 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?
 [/INST] The perpendicular line to the given line has slope $=\frac{-1}{4/3}=-\frac{3}{4}$
Also the perpendicular should pass through the centre of the circle so that the shortest distance between the given line and the given circle is the distance between the given line and the point of intersection of the perpendicular with the circle. SEE THE GRAPH BELOW.
The perpendicular line is $y=-\frac{4}{3}x+\frac{5}{2}$, this intersects with the circle at the point $(\frac{2}{5},\frac{11}{5})$.
The distance between the given line and $(\frac{2}{5},\frac{11}{5})$ is given by:
$$d=\frac{|[3y-4x-20]_{x=2/5,y=11/5}|}{\sqrt{3^2+(-4)^2}}$$
$$=\frac{|\frac{33}{5}-\frac{8}{5}-20|}{\sqrt{25}}=\frac{|-15|}{5}=\frac{15}{5}=3 \text{ units}$$

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