With the error initialized to zero, this algorithm will always step immediately after drawing the first pixel, which will cause that pixel to "stick out". y

With the error initialized to zero, this algorithm will always step immediately after drawing the first pixel, which will cause that pixel to "stick out". y += 1; // y == 1 err += 2*y + 1; // err == 3 x -= 1; // x == radius-1 err -= 2*x + 1; // err == 2-(2*(radius-1)) You could compare the absolute value of the new and old error, or start with err = -(radius-1) instead. And you don't need calculus to come up with the algorithm, just simple high school algebra: (x+1)² - x² = 2x + 1.