# What function could describe this situation on a Cartesian plane?

I have a problem as follows -

I have an arc, with a point on each end. And angle $a$ is how far the arc goes. I need to find the top right point ($P(w, x)$) in relation to the $r$, $P(x, y)$, and $a$ (Note that the x's in either points are not related; I just made a mistake in the diagram). I drew a diagram to better explain. What would describe it? (This is the diagram).

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It would be nicest if the centre of the circle was at the origin.

But you have not said it is. So let us suppose that the centre is at $(a,b)$. Drag the centre to the origin. That moves $(x,y)$ to $(x-a, y-b)=(x',y')$.

Now rotate $(x',y')$ about the origin, counterclockwise, through the angle $a$, which I will call $\theta$ because I am more comfortable with that name.

We end up at a point $(s',t')$ for which there will be a formula soon.

Now undo our earlier dragging, getting $(s'+a,t' +b)$. That is our answer.

Finally, we show how to obtain $(s',t')$ from $(x',y')$. The relevant rotation formula is $$s'=x'\cos\theta-y'\sin\theta,\qquad t'=x'\sin\theta+y'\cos\theta.$$

I hope the many symbols do not cause trouble.

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What do you refer to when you say (a, b) is the center? Do you mean the center as in the origin, or the center as in the vertex where angle θ exists? – Simon Dec 9 '12 at 23:33
I mean the centre of the circle, so the coordinates of the "vertex" where the angle $\theta$ is. I could give an answer purely in terms of $a$, $b$, $x$, and $y$. For example, the first coordinate of the endpoint of the angle is $a+(x-a)\cos\theta -(y-b)\sin\theta$, similar expression for second coordinate. Just thought I would give some indication of where it all comes from. – André Nicolas Dec 9 '12 at 23:37