geometrylinear algebra
Imagine I have two points in 3D space: A (x0, y0, z0) and B (x, y, z). I can draw a line through them. Now what I want is to take a plane that is orthogonal to the line and contains the point A. Let's say point A is a center of a circle that lies in the plane with some radius r. Now, how do I calculate the coordinates of the points that lie on circumference of this circle?
Thank you
You should probably be using Bresenham's Circle Algorithm?
The "distance" between adjacent pixels is either 1 or $\sqrt{2}$, i.e., two adjacent pixels either share a row or column or they are diagonal. This limits how much of the circumference consecutive pixels can consume. So the number of pixels needed, $x$, to complete the circumference of a circles with a radius of $r$ pixels is bounded $\sqrt{2} \pi r \leq x \leq 2 \pi r$, or roughly $4.443 \times r \leq x \leq 6.283\times r$. This gives a first approximation, by which to check the later estimate.
To get more accuracy we see that, by symmetry, only the pixels drawn in the first octant (i.e., $\pi/4$) need to be computed; the remaining can be found by using the same offsets but in different directions. With each consecutive pixel in the first octant progressing upward, there are at least $\lfloor r \sin(\pi/4)\rfloor$ pixels. Multiply this by 8, and you get roughly $5.657\times r$, which is within the earlier bounds.
Let C be the circle located inside the first quadrant. (The logic will be the same if the origin is located at the top left corner.) We let its equation be $F(x, y) = (x – h)^2 + (y – k)^2 - r^2 = 0$.
Initial task == Testing whether the mouse pointer (at T(p, q)) is inside or outside of the circle.
This is easily done by testing whether F(p, q) is greater than or equal to or smaller than 0. The conclusion will be outside/ on/ inside the circle accordingly.
We assume that T(p, q) is inside the circle.
The corresponding point (the one on the circumference) is then (p, m) where m will have 2 values (? and ??).
From the fact that $m – k = \pm \sqrt(r^2 – (p – h)^2)$, and with the help of the diagram, we conclude that $?? = k + \sqrt(r^2 – (p – h)^2)$.
Best Answer
There will be infinite points which lie on the perimeter of the circle. The easiest way to represent a circle in $3D$ would be to take the intersection of a sphere and a plane.
Equation of the sphere would be - $$S : (X-x_o)^2 + (Y-y_o)^2+ (Z-z_o)^2=(x-x_o)^2 + (y-y_o)^2+ (z-z_o)^2$$
Equation of the plane will be - $$P : (x-x_o)(X-x_0) + (y-y_o)(Y-y_0)+ (z-z_o)(Z-z_0) =0 $$
The solutions of these two equations will lie on the circumference of the circle.