Logarithmic Spiral - Is Point on Spiral (cartesian coordinates

Question

Lets Say I have a 3d Cartesian grid. Lets also assume that there are one or more log spirals emanating from the origin on the horizontal plane.

If I then have a point in

Answer 1

here a C++ code drawing any spiral passing where the mouse here (sorry for my English)

int cx = pWin->vue.right / 2;
int cy = pWin->vue.bottom / 2;


double theta_mouse = atan2((double)(pWin->y_mouse - cy),(double)(pWin->x_mouse - cx));
double square_d_mouse = (double)(pWin->y_mouse - cy)*(double)(pWin->y_mouse - cy)+
                        (double)(pWin->x_mouse - cx)*(double)(pWin->x_mouse - cx);
double d_mouse = sqrt(square_d_mouse);
double theta_t = log( d_mouse / 3.0 ) / log( 1.19 );
int x = cx + (3 * cos(theta_mouse)); 
int y = cy + (3 * sin(theta_mouse));

MoveToEx(hdc,x,y,NULL);
for(double theta=0.0;theta < PI2*5.0;theta+=0.1)
{
    double d = pow( 1.19 , theta ) * 3.0;

    x = cx + (d * cos(theta-theta_t+theta_mouse)); 
    y = cy + (d * sin(theta-theta_t+theta_mouse));

    LineTo(hdc,x,y);
}

Ok now the parameter of spiral is 1.19 (slope) and 3.0 (radius at center) Just compare the points where theta is a mutiple of 2 PI = PI2 = 6,283185307179586476925286766559 if any points is near of a non rotated spiral like

x = cx + (d * cos(theta)); 
y = cy + (d * sin(theta));

then your mouse is ON the spiral... I searched this tonight and i googled your past question