Calculating points in an arc [duplicate]

Hi I'm trying to figure out how to calculate the coordinates of a dot at a certain percentage point on an arc. Let's say the dot starts at (800, 300), the half-way points is (400, 0) and the end point is (0, 300). Let's say I want to calculate where the dot will be after traveling 25% of the arc. I'm really clueless where to start on this, thanks for your help.

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marked as duplicate by ja72, martini, hardmath, rogerl, EmilyMar 27 '15 at 1:01

This question was marked as an exact duplicate of an existing question.

Is this an arc of a circle? – Gerry Myerson Sep 29 '12 at 13:10
Well its shape would fit the edge of a circle, but how big that circle would be I have no idea. Imagine the sun rises at (800, 300), the highest point it reaches is (400, 0), and it sets at (0, 300). How would I calculate its coordinates when it's risen 50% (half-way between (800, 300) and (400, 0)) – John Sep 29 '12 at 13:45
There are infinitely many parabolic arcs that go through those three points. Do you have any more information about the function for the arc? – Ben Sep 29 '12 at 13:48
Oh yes I see now >_<. Stupid me, well I will just make a giant circle and chop the top off and use that as my arc then. I can calculate that then since I know the radius, thanks lol. :) – John Sep 29 '12 at 15:36

It appears from the comments that you are happy to take the arc to be an arc of a circle. There is conflicting evidence as to whether you can work out what the circle is, so I'll discuss that. If the radius of the circle is $r$, then, by symmetry, the center of the circle is at $C=(400,r)$. We must have $$(400)^2+(r-300)^2=r^2,$$ an equation which is easy to solve to get $r$.
Now you want the point halfway from $A=(800,300)$ to $B=(400,0)$. You have to find the angle $\theta=\angle ACB$, then you have to bisect it, then you have to find where the bisector meets the arc, that is, you want the point $D$ on the arc for which $\angle ACD=\theta/2$. Come back if you have any trouble carrying out these calculations.