Simple algorithm for drawing filled ellipse in C/C++

Question

On SO, found the following simple algorithm for drawing filled circles:

for(int y=-radius; y<=radius; y++)
    for(int x=-radius; x<=radius; x++)
        i         


        

Answer 1

Replace

x*x+y*y <= radius*radius

with

Axx*x*x + 2*Axy*x*y + Ayy*y*y < radius*radius

where you have three constants, Axx, Axy, Ayy. When Axy=0, the ellipse will have its axes straight horizontal and vertical. Axx=Ayy=1 makes a circle. The bigger Axx, the smaller the width. Similar for Ayy and height. For an arbitrary ellipse tilted at any given angle, it takes a bit of algebra to figure out the constants.

Mathematically Axx, Axy, Ayy are a "tensor" but perhaps you don't want to get into that stuff.

UPDATE - detailed math. I don't think S.O. can make nice math like Math S.E. so this will look crude. You want to draw (or do whatever) with an ellipse in x,y coordinates. The ellipse is tilted. We create an alternative coordinate system x',y' aligned with the ellipse. Clearly, points on the ellipse satisfy

(x'/a)^2 + (y'/b)^2 = 1  

By contemplating some well-chosen random points we see that

x' =  C*x + S*y
y' = -S*x + C*y

where S, C are sin(θ) and cos(θ), θ being the angle of the x' axis w.r.t. the x axis. We can shorten this with notation x = (x,y) and similar for primed, and R a 2x2 matrix involving C and S:

x' = R x

The ellipse equation can be written

T(x') A'' x' = 1

where 'T' to indicates transpose and, dropping '^' to avoid poking everyone in the eyes, so that "a2" really means a^2,

A'' =

1/a2     0  
 0     1/b2

Using x' = Rx we find

T(Rx) A'' Rx = 1

T(x) T(R) A'' R x =1

T(x) A x = 1

where A, the thing you need to know to make your tilted drawing scan line algorithm work, is

A = T(R) A'' R =

C2/a2+S2/b2     SC(1/a2-1/b2)
SC/(1/a2-1/b2)  S2/a2 + C2/b2    

Multiply these by x and y according to T(x)Ax and you've got it.