I need to solve problem very similar to Generating polygon representing rough 100km circle around latitude/longitude point using Python? but subtly different.

I need a shape that will be a circle once projected in Web Mercator - not in reality.

Working answer ( https://gis.stackexchange.com/a/268251/45061 ) is not generalizable, as it relies on https://github.com/jwass/geog that will not allow doing what I am trying to do.

  • 4
    A geometric object which is a circle when presented in Web Mercator is of very limited use ("How not to generate geometries" seminar is all I can think of). The easiest way to generate is a simple buffer in Web Mercator, then deproject. But first you need to select a GIS software stack.
    – Vince
    Commented Mar 8, 2021 at 20:28
  • @Vince I know that it is fairly unusual. If anyone worries about using it for anything serious, it is for generating boundaries of laser-cut decoration plates on scales small enough that usual Web Mercator deficiencies are not relevant (city centers - so remembering to keep the same scale should be sufficient). Also, as it is for decorations most of Web Mercator issues are not really relevant. Commented Mar 8, 2021 at 22:48
  • Web Mercator's deficiencies as a map projection are so numerous that I consider paper with ink jet ink an unconscionable waste of materials in map production. I can't imagine using any more exotic material being used. If the print cost is measured in dollars, a local Albers Equal Area or Lambert Conformal would be a more worthy projection.
    – Vince
    Commented Mar 9, 2021 at 12:04

1 Answer 1


Despite it being an abomination in so many ways, Pseudo/Spherical (Web) Mercator is a projection after all, so basic math on the Cartesian plane applies:

import math

R = 6378137.0

def LL2SM(ll):
  return [math.radians(ll[0]) * R, math.log(math.tan(math.pi / 4 + math.radians(ll[1]) / 2)) * R]

def SM2LL(xy):
  return [math.degrees(xy[0] / R), math.degrees(2 * math.atan(math.exp(xy[1] / R)) - math.pi / 2.0)]

def VERTEX(cxy, a, r):
  return [cxy[0] + r * math.cos(math.radians(a)), cxy[1] + r * math.sin(math.radians(a))]

def main():
  vertices = 32

  center = [13.0, 52.0]
  radius = 100000

  circle = []

  _angle = 360.0/vertices
  for vertex in range(0, vertices):
    sm = LL2SM(center)
    cv = VERTEX(sm, vertex*_angle, radius)
    ll = SM2LL(cv)


  # do sth. with your circle

if __name__ == "__main__":

This POC script

  1. transforms a pair of Longitude/Latitude (center) into EPSG:3857 coordinates (LL2SM())
  2. generates vertices amount of points on a circle with a given radius (in meter) around center, using its parametric equations (VERTEX())
  3. transforms each vertex' coordinates back to Longitude/Latitude (SM2LL())

circle will then hold coordinate arrays ([Longitude, Latitude]) which, when transformed again to EPSG:3857, will form a circle around center.

Obviously, you can skip transforming back to Longitude/Latitude if you want.

You haven't specified any software environment other than Python, so I leave translating this into the framework of your choice to you; this includes generating an actual Polygon, for which there likely are built-in functions.


# do sth. with your circle
wkt_polygon = ''.join(['POLYGON((', ','.join([' '.join(map(str, ll)) for ll in circle + [circle[0]]]), '))'])


will output a Polygon as valid Well Known Text representation.

  • Note that this is obviously the basic and manual math version of the general workflow (project center -> create buffer -> deproject buffer), which should be covered in all spatial frameworks. Good to know the math, though. Keep it simple once in a while.
    – geozelot
    Commented Mar 16, 2021 at 11:02

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