Please excuse my poorly drawn doodle here, I'm almost inept at drawing.
I'm attempting to compute i2, j2, x2, y2.
Knowns: x1, y1, xk, yk, i1, j1, the arc is circular
Constraints:
resulting arc is circular
cartesian co-ordinate system
intermediate calculations can be in any co-ord system that makes sense
problem may be arbitrarily rotated, and so can't assume y2=j2 or i2=xk, etc.
x2 and y2 are not known, but the arc length from x1,y1 to x2,y2 is an arbitrary but fixed, known distance, say 100 units.
I suppose where I'm getting most confused is that there's an arc involved. If this were triangles even I could solve this problem. I just don't know my properties of arcs well enough to make valid assumptions.
Is it a valid approach to manipulate the arc length formula to supply x2,y2 and then use similar triangles to solve back for i2,j2?
Can we assume that the line formed by x2,y2-i2,j2 is always parallel to x1,y1 to i1,j1 regardless of rotation because of the constraints (ie since the initial and resulting arcs are circular, do these lines not form a right angle wtih the line (i1,j1), (xk,yk) ) ?
Cannot make this assumption, counterexample let the angle = 180 degrees.
This is not a homework problem, it's something I need to understand for a project I'm involved with.
Any help is greatly appreciated.
I'm going to make some assumptions based on the comments. We have a circular arc, which is a part of a circle with center $(i,j)$, and the endpoints of the arc are $(x_1,y_1)$ and $(x_k,y_k)$, and we want to remove a piece of length $\ell$ from the $(x_1,y_1)$ end of the arc and find the new endpoint of the arc $(x_2,y_2)$.
From the given endpoints and center, we can find the radius $r$ of the arc (find the distance between either endpoint and the center). The arc we're removing has length $\ell$, so it has measure $\theta=\frac{\ell}{r}$ (if $\theta$ is the measure of an arc, the length of the arc is $r\theta$).
We now want to rotate $(x_1,y_1)$ by $\theta$ along the given arc, about $(i,j)$. I'm going to assume this is a clockwise rotation (which may or may not be correct). The result of this rotation is $$\begin{align}(x_2,y_2)&=(i+(x_1-i)\cos\theta+(y-i)\sin\theta,j+(y-j)\cos\theta+(i-x)\sin\theta)\\&=(i+(x_1-i)\cos\frac{\ell}{r}+(y-i)\sin\frac{\ell}{r},j+(y-j)\cos\frac{\ell}{r}+(i-x)\sin\frac{\ell}{r})\end{align}.$$