If you simply want to find out if a point lies within a circle you could measure the distance from the circle center to the point. If the distance to the point is less than the radius of the circle then the point is within the circle.
Func GetDistance($fX1, $fY1, $fX2, $fY2)
Return Sqrt(($fX1 - $fX2) ^ 2 + ($fY1 - $fY2) ^ 2)
EndFunc
If the point is within the circle you could then get the angle from the circle center to the point. Knowing the distance and the angle would tell you exactly which part of the circle the pixel is in.
Func GetAngle($fX1, $fY1, $fX2, $fY2)
Return ((($fY2 <= $fY1) * 180) - (ATan(($fX1 - $fX2) / ($fY1 - $fY2)) * 57.295779) + 90)
EndFunc
As far as getting all the pixel coordinates of a circle, I can’t think of an efficient way to do it. This would work but is very slow...
Func GetCircleCoordinates($fCircCentX, $fCircCentY, $fCircRadius)
Local $aCoordinates[Abs(($fCircRadius * 2) ^ 2) + 1][2]
For $iY = Abs($fCircCentY - $fCircRadius) To Abs($fCircCentY + $fCircRadius)
For $iX = Abs($fCircCentX - $fCircRadius) To Abs($fCircCentX + $fCircRadius)
If (GetDistance($fCircCentX, $fCircCentY, $iX, $iY) < $fCircRadius) Then
$aCoordinates[0][0] += 1
$aCoordinates[$aCoordinates[0][0]][0] = $iX
$aCoordinates[$aCoordinates[0][0]][1] = $iY
EndIf
Next
Next
ReDim $aCoordinates[$aCoordinates[0][0] + 1][2]
Return $aCoordinates
EndFunc