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 ?

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- Aug 2nd 2012, 02:12 PMRobertthenewtHow to move point 'p' through a tangent.
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 ? - Aug 2nd 2012, 02:26 PMemakarovRe: How to move point 'p' through a tangent.
What does it mean to move a point through a tangent?

- Aug 2nd 2012, 02:49 PMskeeterRe: How to move point 'p' through a tangent.
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. - Aug 2nd 2012, 03:16 PMRobertthenewtRe: How to move point 'p' through a tangent.
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. - Aug 2nd 2012, 05:25 PMskeeterRe: How to move point 'p' through a tangent.
you need to make a sketch ...