Welcome to Math.SE! Here is my sketch for the case of clockwise direction: initial point is $A$, the endpoint you want is $B$, and $C$ is the radius of the circle on which the arc lies.
I will use polar angles: you can see the description in Wikipedia. The formulas for conversion from polar to Cartesian (on the same wiki article) will also be used.
On the picture, $\theta$ is the polar angle of the direction in which the curve departs from point $A$. Since any radius of a circle is perpendicular to the circle, the polar angle of the vector $\vec{CA}$ is $\theta+\pi/2$. The length of $\vec{CA}$ is $r$. Therefore, its Cartesian coordinates are
$$\vec {CA} = ( r\cos(\theta+\pi/2), r\sin(\theta+\pi/2)) = ( -r\sin(\theta), r\cos(\theta)) $$
Next, we need the coordinates of the vector $\vec{CB}$. Its length is also $r$. Since $\angle ACB$ is $L/r$ radian, the polar angle of $\vec{CB}$ is $\theta+\pi/2-L/r$. Convert to Cartesian:
$$\vec {CB} = ( r\cos(\theta+\pi/2-L/r), r\sin(\theta+\pi/2-L/r)) = ( -r\sin(\theta-L/r), r\cos(\theta-L/r)) $$
Finally, $\vec{AB}=\vec{CB}-\vec{CA}$, which yields
$$\boxed{\vec {AB} = ( -r\sin(\theta-L/r)+r\sin(\theta), r\cos(\theta-L/r)-r\cos(\theta)) } $$
These can be rewritten using some trigonometric identities, but I don't think it would win anything. As a sanity check, consider what happens when $L=0$: the vector is zero, hence $B$ is the same as $A$. As an aside, if $r\to \infty$ the curve becomes a straight line segment, but figuring out the limit is an exercise in calculus. :-)
If the curve bends counterclockwise, the signs will be different in a few places. Namely, the polar angle of $\vec{CA}$ will be $\theta-\pi/2$, hence
$$\vec {CA} = ( r\sin(\theta), -r\cos(\theta)) $$
The polar angle of $\vec{CB}$ will be $\theta-\pi/2+L/r$, hence
$$\vec {CB} = ( r\sin(\theta+L/r), -r\cos(\theta+L/r)) $$
The conclusion in this case is
$$\boxed{\vec {AB} = ( r\sin(\theta+L/r)-r\sin(\theta), -r\cos(\theta+L/r)+r\cos(\theta))}$$
Later: a simpler solution for the case when $C$ is given. First, calculate the vector $\vec{CA}$ and convert it to polar coordinates using these formulas. Then either increase or decrease the angle by $L/r$, depending on counterclockwise/clockwise choice.
Since you wanted JavaScript, I made a jsfiddle and also copied the code below. The parameters are coordinates of A and C, as well as length of the arc and the direction. The radius $r$ is calculated within the function.
function findB(Ax, Ay, Cx, Cy, L, clockwise) {
var r = Math.sqrt(Math.pow(Ax - Cx, 2) + Math.pow(Ay - Cy, 2));
var angle = Math.atan2(Ay - Cy, Ax - Cx);
if (clockwise) {
angle = angle - L / r;
}
else {
angle = angle + L / r;
}
var Bx = Cx + r * Math.cos(angle);
var By = Cy + r * Math.sin(angle);
return [Bx, By];
}
document.write(findB(0, 1, 1, 0, 1, true));
Given a quadratic Bezier curve
$$
b_2(p_0,p_1,p_2,t) = (1-t)((1-t)p_0+t p_1)+t((1-t)p_1+t p_2),\ \ 0\le t\le 1
$$
we have the control point as $p_1$ now the question is. Where locate $p_1$ such that the spline passes by $p_0,q,p_2$?
we have for a suitable set $\{t^*, p_1^*\}$ that
$$
b_2(p_0,p_1^*,p_2,t^*) = q
$$
so fixing a $0\le t^* \le 1$ and giving a desired $q$ we can determine the control point $p_1^*(t^*)$ by solving
$$
(1-t^*)((1-t^*)p_0+t^* p_1^*)+t^*((1-t^*)p_1^*+t^* p_2) = q
$$
Attached a plot showing the case
$$
q = (0.8,0.6) \ \ \mbox{red}\\
p_0 = (0,2) \ \ \mbox{red}\\
p_1^* = (1.66667,0.416667) \ \ \mbox{green}\\
p_2 = (0,-2) \ \ \mbox{red}\\
t^* = 0.4
$$
NOTE
Be aware that $p_1^* = p_1^*(t^*)$
Best Answer
if $(a,b)$ is the cordinates of the known point. then unknown point is $(a+r\cos \theta,b+r\sin \theta)$ where $r$ is the known length and $\theta$ the angle.