code: def places_within_radius(
          self, place=None, latitude=None, longitude=None, radius=0, **kwargs
      ):
          
          kwargs['withdist'] = True
          kwargs['withcoord'] = True
          kwargs['withhash'] = False
          kwargs.setdefault('sort', 'ASC')
          unit = kwargs.setdefault('unit', 'km')
  
          # Make the query
          if place is not None:
              response = self.redis.georadiusbymember(
                  self.key, self._pickle(place), radius, **kwargs
              )
          elif (latitude is not None) and (longitude is not None):
              response = self.redis.georadius(
                  self.key, longitude, latitude, radius, **kwargs
              )
          else:
              raise ValueError(
                  'Must specify place, or both latitude and longitude'
              )
  
          # Assemble the result
          ret = []
          for item in response:
              ret.append(
                  {
                      'place': self._unpickle(item[0]),
                      'distance': item[1],
                      'unit': unit,
                      'latitude': item[2][1],
                      'longitude': item[2][0],
                  }
              )
  
          return ret
docs: Return descriptions of the places stored in the collection that are
          within the circle specified by the given location and radius.
          A list of dicts will be returned.
  
          The center of the circle can be specified by the identifier of another
          place in the collection with the *place* keyword argument.
          Or, it can be specified by using both the *latitude* and *longitude*
          keyword arguments.
  
          By default the *radius* is given in kilometers, but you may also set
          the *unit* keyword argument to ``'m'``, ``'mi'``, or ``'ft'``.
  
          Limit the number of results returned with the *count* keyword argument.
  
          Change the sorted order by setting the *sort* keyword argument to
          ``b'DESC'``.