If I have point p defined as P(x,y) where :
Sin(theta) = y/r and Cos(theta) = x/r.
How can I move point p through a tangent so that 'P' is
then twice the distance from the origin ?
if I interpret your post correctly, $\displaystyle P$ is a point $\displaystyle (x,y)$ on a circle of radius $\displaystyle r$.
let the line tangent to the circle at point P be $\displaystyle \overline{PP'}$ where the length of segment $\displaystyle \overline{PP'}$ is $\displaystyle r\sqrt{3}$ ...
the points $\displaystyle P$ , $\displaystyle P'$ and the origin form a 30-60-90 triangle and point P' will be a distance 2r from the origin.
Ah, almost. Except if the triangle was looked at with the top being the origin and below that P and to the right of that P1. what I was hoping to find a solution for is that this triangle could be at any angle on the circle but the point I would be interested in would be a point projected beneath p by a vector length of size n. So, that the origin and P and P2 would then line up. An extrusion along a path might be a better term to use.