<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] By substituting the equation of the line into the equation of the circle you are looking for points where the line intersects the circle. The fact that the resulting equation for $x$ has no real roots means that the line does not intersect the circle i.e. the shortest distance from the line to the circle will be greater than the radius of the circle, which is $2$ units.
You can find the shortest distance from the line to the circle as follows:
(1) Note that the product of the gradients of perpendicular lines is $-1$, so the general equation of a line perpendicular to the given line is
$y = - \frac 3 4 x + m$
(2) The line with gradient $-\frac 3 4$ that passes through the centre of the circle at $(2,1)$ is
$y = - \frac 3 4 x + \frac 5 2$
(3) This line interects the original line at a point where
$\frac 4 3 x + \frac {20} 3 = - \frac 3 4 x + \frac 5 2 \\
\Rightarrow 16x + 80 = -9x + 30 \\
\Rightarrow 25x = -50 \\
\Rightarrow x=-2 \text{ and } y=4$
(4) Find the distance between $(-2, 4)$ and the centre of the circle $(2,1)$. Then subtract the radius of the circle from this distance - this is the shortest distance from the line to the circle.
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