Thread: How 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 ?

What does it mean to move a point through a tangent?

Originally Posted by Robertthenewt

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, is a point on a circle of radius .

let the line tangent to the circle at point P be where the length of segment is ...

the points , 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.

you need to make a sketch ...