code: def _call_api_single_related_resource(self, resource, full_resource_url,
                                            method_name, **kwargs):
          
          url = full_resource_url
          params = {
              'headers': self.get_http_headers(
                  resource.Meta.name, method_name, **kwargs),
              'url': url
          }
          prepared_request = self.prepare_http_request(
              'GET', params, **kwargs)
          response = self.session.send(prepared_request)
          return self._handle_response(
              response, resource.Meta.valid_status_codes, resource)
docs: For HypermediaResource - make an API call to a known URL