# Recomputing arc center

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.

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Next time you may want to use GeoGebra for drawing such a picture :) –  t.b. Jul 12 '11 at 13:29
@Theo Buehler Looking into it now... but it's only useful if it can help me derive formulas. –  Stephen Jul 12 '11 at 13:46
@Theo Buehler It's not explicitly stated, and I don't believe we can make that assumption... I think the conditions all hold if the angle were 180 degrees (for example)... –  Stephen Jul 12 '11 at 13:51
I'm sorry but I'm still confused. Let me make a guess: Are you trying to divide a circular arc from $(x_1,y_1)$ to $(x_k,y_k)$ with center $(i_1,j_1)$ into $(k-1)$ (or maybe $k$) equal pieces and want to know the coordinates $(x_2,y_2)$ (and maybe $(x_3,y_3)$, etc later on)? @Isaac: ping! –  t.b. Jul 12 '11 at 16:15
Yes, of course you're correct ... a great light bulb has dimly flickered on. Somewhere in the description of the problem I was under the impression that i and j (as the inputs to the computer functions) had to change in a particular was and this kept me from seeing what you were saying. THANKS SO MUCH for everyone's patience. –  Stephen Jul 12 '11 at 17:20

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}.
@Stephen: For a counterclockwise rotation, you could just change $\theta$ to $-\theta$, or change the $y-i$ in front of $\sin$ in the $x$ term to $i-y$ and the $i-x$ in front of $\sin$ in the $y$ term to $x-i$. ($\cos(-\theta)=\cos\theta$, so the cosine terms are unaffected, but $\sin(-\theta)=-\sin\theta$, so the sign of the sine terms changes, which is equivalent to swapping the order of the subtraction in their coefficients.) –  Isaac Jul 12 '11 at 18:01