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?
 A: 
According to the figure, you need the distance $AB$. 
The line $3x+4y=10$ is the line passing through $(2,1)$ and perpendicular to the line $3y=4x+20$
A: 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.
A: The line perpendicular to the line given in the question and passing through the centre will help us find the perpendicular distance.
We approach the problem by first finding out the perpendicular distance of the given line from the centre using this formula (check under Cartesian coordinates). Applying the formula we get distance between line and center to be:
$$\left|\dfrac{3(1)-4(2)-20}{5}\right|=5$$
Now for the shortest distance between the given line and circle we simply subtract the radius of the circle giving us the answer of $\boxed{3}$
What you're trying to do is to intersect the line and circle. If you did get a real solution to that it would imply the shortest distance is $0$! 
A: 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:
$$d=\frac{|[3y-4x-20]_{x=2/5,y=11/5}|}{\sqrt{3^2+(-4)^2}}$$
$$=\frac{|\frac{33}{5}-\frac{8}{5}-20|}{\sqrt{25}}=\frac{|-15|}{5}=\frac{15}{5}=3 \text{ units}$$

A: 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:
$5-2=\boxed{3}$
A: 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!
A: 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 $2$.
