Iterating over a solid circle of pixels

Given a 2d array of pixel values, is it possible to iterate over just the pixels in a circle in a way that is less costly than simply iterating over a square and checking if a value is within the circle?

I need to get the values of pixels within a circle given center coordinates and the radius. Ideally I would ONLY iterate over the pixels in the circle as the leave number of iterations possible is the least costly for time.

The only solution I have found so far is to iterate over a square(shown in green below) and for each pixel, check if its inside the circle using pythagoras before performing the logic I wish to do.

Is there a way to only iterate over the solid circle of pixels alone?

(Int the image below, the desired pixels are highlighted in RED and the containing square is in GREEN)

thank you in advance for any help and resources offered • Well, at the very least, you could calculate the first and last pixel on each row and iterate between them. Symmetry about the two co-ordinate axes should also help: if the centre is at $(a,b)$ and you know that $(a-x,b-y)$ is in the circle, you know that $(a+x,b-y)$, $(a-x,b+y)$ and $(a+x,b+y)$ are all in it, too. – David Richerby Feb 8 '18 at 18:24
• Have you tried to simply calculate starting point of the row, mirroring it to find end (also the lower half of the circle)? You have center, radius and y, solve for x? Or simply check Bresenham Circle algorithm, filling (reading) inside row by row (scanline). – Evil Feb 8 '18 at 18:24
• @Evil Haha! I beat you by six seconds, even though my comment is twice as verbose as yours! ;-) – David Richerby Feb 8 '18 at 18:27
• @DavidRicherby that was true, now it is only 3/2 times more verbose ;) I was looking for link with a picture of octant symmetry Bresenham style and midpoint standard. – Evil Feb 8 '18 at 18:29
• Thanks @DavidRicherby and @Evil! In case you're interested, I have implemented both my original Pythagoras on a square method, and Bresenham's Algorithm in JavaScript. I have found the Pythagoras method to be 2.8x faster for my use case :D – drm18272 Feb 10 '18 at 20:42