Creating a circle in actual meters

The following is only valid for differential elements of a direct Mercator projection on a unit sphere:

• In the direction of parallels, deformation modulus is constant. And a parallel circumference measures 2 π cos(φ) on the sphere surface and 2 π on the projected image.

• We know that it is conform, so we will save ourselves from calculating the deformation in the direction of the meridians, which respect the orthogonality with the parallels on both surfaces. (The projection is analytically conform, the transformation in the direction of the meridians is the one that fulfills that the deformation modulus in that direction is the same as in the parallels one.)

• So being conform, let's say the longitudinal Mercator modulus of deformation at a point is 1 / cos(φ) in any direction.

Although this is not true for a Web Mercator projection of a non-spherical objective surface, it is nevertheless quite close.

So if the tolerance of your work allows it, one way to reach your goal is to divide the actual meters by the cosine of the latitude of the center of the circle, and use that measurement as the radius.

One option would be to create a circular polygon with the correct radius. The function below makes a 32 vertex circular polygon centered on a point:

function drawCircle(point, radius, dir) { 
  var d2r = Math.PI / 180;   // degrees to radians 
  var r2d = 180 / Math.PI;   // radians to degrees 
  var earthsradius = 6378137.0; // 6378137.0 is the radius of the earth in meters

  var points = 32; 

  // find the raidus in lat/lon 
  var rlat = (radius / earthsradius) * r2d; 
  var rlng = rlat / Math.cos(point[1] * d2r); 

  var extp = new Array(); 
  if (dir==1)  {var start=0;var end=points+1} // one extra here makes sure we connect the
  else     {var start=points+1;var end=0}
  for (var i=start; (dir==1 ? i < end : i > end); i=i+dir)  
  { 
    var theta = Math.PI * (i / (points/2)); 
    ey = point[0] + (rlng * Math.cos(theta)); // center a + radius x * cos(theta) 
    ex = point[1] + (rlat * Math.sin(theta)); // center b + radius y * sin(theta) 
    extp.push([ey, ex]); 
  } 
  return extp;
}

live example (red circle from the function, blue circle is the OpenLayers Circle)