Quickest way to find a point on a circumference Given the image below, A is the centre of the circle, B is a point on the circumference and AC and DB lie on parallel lines.
Knowing A, C, D and the radius of the circumference, I need to find the quickest way (in term of # of calculations) to find B. 

EDIT:
The position of C can vary, AC can also be parallel to x or y axis. This is for a program I'm making, I need to handle also corner cases in terms of calculations where the slope can be infinity (computers don't work well with infinity). In this context I found that using trigonometric functions I can achieve better results, but in this particular case I've come to no good solutions.
 A: Let $A:={\bf 0}$, $C:={\bf c}$, $D:={\bf d}$, and $B:={\bf b}$. Then you want ${\bf b}={\bf d}+\lambda{\bf c}$ for some $\lambda>0$ and $|{\bf b}|^2=r^2$, where the radius $r>0$ is assumed to be given. Therefore
$$|{\bf d}|^2 + 2\lambda\langle{\bf d},{\bf c}\rangle +\lambda^2|{\bf c}|^2=r^2\ ,$$
which can be solved for $\lambda$ from the given data. Then ${\bf b}={\bf d}+\lambda{\bf c}$.
A: You can't (unless you are assuming these lines are parallel, which I will assume because your diagram suggests it.)
Compute the slope of the line through $A$ and $C$.
Use variables for the coordinates of $B$ and fill out the formula for the slope of the line through $B$ and $D$, using the fact that it's the same slope as the parallel line.

Rereading it again, I should also clarify that I'm assuming $A$ is the center, and you need to know what the radius of the circle is. Otherwise the picture is too vague.
A: Sounds like you want a programming-oriented solution, rather than a mathematical one. So, let's assume that you have (or can write) two standard functions:
Dot(u,v) -- computes the dot product of two 2D vectors $u$ and $v$
QuadraticRoots(a,b,c) -- computes the roots of the quadratic equation $ax^2+bx+c=0$
Then, following Christian Blattner's answer, we first calculate a vector $U = C-A$.
We want to find $t$ such that $D+tU$ lies on the circle. But this means that $\lVert D+tU-A\rVert = r$, or $(D+tU-A)\cdot(D+tU-A) = r^2$.
Multiplying out the dot product, this means that:
$$(U\cdot U)t^2  + 2U\cdot(D-A)t + (D-A)\cdot(D-A) = r^2$$
This is a quadratic equation that you can solve for $t$. The pseudocode is:
U = C - A
V = D - A
a = Dot(U,U)
b = 2*Dot(U,V)
c = Dot(V,V) - r*r
(t1,t2) = QuadraticRoots(a,b,c)
If t1 > 0, then t = t1, else t = t2
B = D + t*U

If $A$ and $D$ are fixed, and only $B$ is varying, then you should precompute $V$ and $c$, of course, to save time.
A: Assuming that $A$ is the center of the circle, that $\overline{AC}$ and $\overline{BD}$ are parallel, and that you know the radius of the circle ($r$), and that $\overline{AC}$ and $\overline{BD}$ are parallel (If you don't know the center or the radius of the circle, you have no way of confirming that $B$ is on its circumference, and we can't tell anything about $\overline{BD}$ otherwise):
Compute the slope of the line through $\overline{AC}$ ($m$).
Use point-slope form of a line to define a line for $\overline{BD}$: $ y - y_D = m ( x - x_D )$
From here, you need to find a point on $\overline{BD}$ with distance $r$ from $A$.
You have the equation of a circle: $ r^2 = \sqrt{ (x - x_A)^2 + (y - y_A)^2 }$ 
Multiply the equation for the circle and isolate $y^2$ on one side. Isolate $y$ in the equation for the line, and then square that, so you have $y^2$ on one side by itself. That done, you can equate the two equations together, and then isolate y to form a quadratic equation.
You'll get two answers for the x-coordinate, so you'll need to do a sanity check to make sure you use the right one. (The other one will be the intersection of $\overline{BD}$ and the circle on the other side of $D$ from $B$.) Plug it into your line equation and you'll get the other part of your coordinate.
