Check if line intersects with circles perimeter Say I have this circle here. 
What would be an algorithm to check if lines like the green or blue lines intersect the edge of the circle, but not the red line.
So lets say
if(amountOfPointsHit(line, circle) > 1) then return true
else return false


I only know the start and finish points of the lines, and not where the line intersects the circle. So the pseudo code would more be like 
if(pointBetweenXYHit(x1, y1, x2, y2, circle) > 1) then return true
else return false

Here is the code I came up with based on the answers...
/**
*@param l1  Line point 1, containing latitude and longitude
*@param l2  Line point 2, containing latitude and longitude
*@param c   Center of circle, containing latitude and longitud
*@param r   Radius of the circle
**/
Maps.ui.inCircle = function(l1, l2, c, r){
    var a = l1.lat() - l2.lat()
    var b = l1.lng() = l2.lng()
    var x = Math.sqrt(a*a + b*b)
    return (Math.abs((c.lat() - l1.lat()) * (l2.lng() - l1.lng()) - 
           (c.lng() -  l1.lng()) * (l2.lat() - l1.lat())) / x <= r);
}

pseudo code for that ^^ is 
method inCircle (line1point, line2point, center, radius)
    let l1 = line1point
    let l2 = line2point
    let c = center
    let r = radius
    x is length between points line1point and line2point
    return true if ((c.x - l1.x) * (l2.y - l1.y) - (c.y - l1.y) * (l2.x - l1.x)) / x) 
    is less then or eqaul to r

 A: Let $A = (A_x,A_y)$ and $B = (B_x,B_y)$ be the end points of a line segment. Then all points of the line are $A + t (B-A)$ for $0 < t < 1$.
Let $C = (C_x,C_y)$ be the center of the circle and $R$ its radius. Then all points of the circle are $(x,y)$ such that $(x-C_x)^2 + (y-C_y)^2 = R^2$ by Pythagoras theorem.
By moving everything we can have $C = (0,0)$, this makes calculations a bit simpler. The equation of the circle becomes $x^2 + y^2 = R^2$.
As for number of points of intersection: there will be either 0 - no intersection, 1 - it is a tangent line or 2 - it goes right through the circle.
The points of intersection must satisfy both equations simultaneous. $(x,y)$ is a point of intersection if $x^2 + y^2 = R^2$ and $(x,y) = A + t (B-A)$ for some $0 < t < 1$.
We can split $(x,y) = A + t (B-A)$ into components:


*

*$x = A_x + t (B_x - A_x)$

*$y = A_y + t (B_y - A_y)$
and put this into the circle equation
$$(A_x + t (B_x - A_x))^2 + (A_y + t (B_y - A_y))^2 = R^2$$
multiply it out
$$[A_x^2 + A_y^2 - R^2] + 2 [A_x (B_x - A_x) + A_y (B_y - A_y)] t + [(B_x - A_x)^2 + (B_y - A_y)^2] t^2 = 0$$
this is a simple quadratic equation, you can use the discriminant ("$b^2 - 4ac > 0$") to check if there are two real values of $t$, then you must check if they are between 0 and 1.
// parameters: ax ay bx by cx cy r
ax -= cx;
ay -= cy;
bx -= cx;
by -= cy;
a = (bx - ax)^2 + (by - ay)^2;
b = 2*(ax*(bx - ax) + ay*(by - ay));
c = ax^2 + ay^2 - r^2;
disc = b^2 - 4*a*c;
if(disc <= 0) return false;
sqrtdisc = sqrt(disc);
t1 = (-b + sqrtdisc)/(2*a);
t2 = (-b - sqrtdisc)/(2*a);
if((0 < t1 && t1 < 1) || (0 < t2 && t2 < 1)) return true;
return false;

A: $\newcommand{\norm}[1]{\lVert #1 \rVert}
\newcommand\inner[2]{\langle #1, #2 \rangle}$
Frequently in computer graphics, if you work through the math in terms of vectors instead of coordinates, it makes the math simpler and works for any dimension. Say $v_1$ and $v_2$ are the endpoints of the line segment. Let $v = v_2 - v_1$ be a vector and WLOG let $C$ be a circle (or sphere in higher dimensions) centered at the origin with radius $r$. Then points on the line segment are given by $x = v_1 + t v$ for $0 \le t \le 1$.
We know points on the sphere are given by $\norm{x} = r$. Now we can simply solve $\norm{v_1 + tv}^2 = r^2$ and if there are real solutions we can check the values of $t$ to see if the intersections are within the line segment.
This gives us the quadratic equation in $t$
$$\norm{v}^2 t^2 + 2 \inner{v_1}{v} t + (\norm{v_1}^2 - r^2) = 0$$
We get real solutions for $t$ if discriminant $\Delta = 4 \inner{v_1}{v}^2 - 4\norm{v}^2 (\norm{v_1}^2 - r^2) \ge 0$, and the line is tangent if $\Delta = 0$. We can check if the intersections happen within the line segment by checking if
$$
t = \frac{-2 \inner{v_1}{v} \pm \sqrt \Delta} {2 \norm{v}^2} 
= \frac{- \inner{v_1}{v} \pm \sqrt {\Delta/4}}{\norm{v}^2}
$$
is between $0$ and $1$. These can be simplified in terms of coordinates for special cases of 2D and 3D and should be simplified in code for computational efficiency. For example, to just check if there are intersections for an infinite line, we only need to check if $r^2 ((x_2 - x_1)^2 + (y_2 - y_1)^2) \ge x_1 y_2 - x_2 y_1$. No division or squareroots in this computation!
A: A line segment intersects the edge of the circle if the distance between the center and the line segment is less than or equal to the radius and at least one end of the line segment is outside the circle.
A: You can find the shortest distance from a point to a line using the formula
$$\operatorname{distance}(ax+by+c=0, (x_0, y_0)) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}. $$
Put $(x_0,y_0)$ = center of circle. If this distance is smaller (or equal) than radius of circle, then your line and circle intersects.
Since you know start  point $(x_1,y_1) $ and end point $(x_2, y_2) $, you can get the equation of line using formula
$$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x-x_1)$$
Simplifying, we get
$$ (x_2 - x_1) y + (y_1 - y_2)x +(x_1-x_2)y_1 + x_1(y_2-y_1) = 0$$
It would be nice to store $a = y_1 - y_2, b = x_2 - x_1, c =  (x_1-x_2)y_1 + x_1(y_2-y_1)$
It would be something like
return (Math.abs((l2.lat() - l1.lat())*c.lng() +  c.lat()*(l1.lng() -     
       l2.lng()) + (l1.lat() - l2.lat())*l1.lng() +
       (l1.lng() - l2.lng())*l1.lat())/ Math.sqrt((l2.lat() - l1.lat())^2 +
       (l1.lng() - l2.lng())^2) <= r)

something like 
$$ \frac{\left | (x_2 - x_1)x_0 + (y_1 - y_2)y_0 + (x_1-x_2)y_1 + x_1(y_2-y_1) \right |}{\sqrt{(x_2 - x_1)^2 + (y_1 - y_2)^2}} \le  r$$
