Calculate a vector from the center of a square to edge based on radius

Question

Given a square (described by x, y, width, height) and an angle (in radians) I need to calculate a vector that originates at the squares centre and terminates at the point th

Answer 1

The vector will be center + (cos(angle), sin(angle))*magnitude. Given that you want to intersect this with a square, you need to determine magnitude. You can get that with a square with:

float abs_cos_angle= fabs(cos(angle));
float abs_sin_angle= fabs(sin(angle));
if (width/2/abs_cos_angle <= height/2/abs_sin_angle)
{
    magnitude= fabs(width/2/abs_cos_angle);
}
else
{
    magnitude= height/2/abs_sin_angle;
}

However, cos(angle) or sin(angle) could be zero, so you should cross multiply that out to get:

float abs_cos_angle= fabs(cos(angle));
float abs_sin_angle= fabs(sin(angle));
if (width/2*abs_sin_angle <= height/2*abs_cos_angle)
{
    magnitude= width/2/abs_cos_angle;
}
else
{
    magnitude= height/2/abs_sin_angle;
}

And you can trivially get the end point from that.

EDIT: Here's a snippet you can drop in place to verify this works with the currently accepted answer:

    double magnitude;
    double abs_cos_angle= fabs(cos(angle));
    double abs_sin_angle= fabs(sin(angle));
    if (width/2*abs_sin_angle <= height/2*abs_cos_angle)
    {
        magnitude= width/2/abs_cos_angle;
    }
    else
    {
        magnitude= height/2/abs_sin_angle;
    }

    double check_x= x + cos(angle)*magnitude;
    double check_y= y + sin(angle)*magnitude;

    printf("  a = %d deg: x = %lf; y = %lf\n",(int)(angle/pi*180),check_x,check_y);

Clearly this is applies to an axis aligned rectangle. You can do something similar by finding the closest intersection between the testing vector and every edge in a polygon. (You can optimize that further, but that's left as an exercise to the reader.)