消除平均数值的子查询

Quest

查询选择以“Vancouver”开头且距离以“Vancouver”开头的所有位置的中心 5 分钟范围内的所有点。例如，Vancouver South Fraser、Vancouver Fairview 和Vancouver Ballantree Place W 的纬度和经度与其平均纬度和经度相差不超过 5 分钟。纬度和经度存储为 (4915, 12311) 整数对（即 49.15'N 和 123.11'W）。

SQL Code

下面的令人厌恶的 SQL 就达到了目的：

SELECT
  NAME
FROM
 STATION
WHERE
      DISTRICT_ID = '110'
  AND NAME LIKE 'Vancouver%'
  AND LATITUDE BETWEEN
    (SELECT round((min(LATITUDE) + max(LATITUDE)) / 2)-5 FROM STATION WHERE DISTRICT_ID = '110' AND NAME LIKE 'Vancouver%')
    and
    (SELECT round((min(LATITUDE) + max(LATITUDE)) / 2)+5 FROM STATION WHERE DISTRICT_ID = '110' AND NAME LIKE 'Vancouver%')
  AND LONGITUDE BETWEEN
    (SELECT round((min(LONGITUDE) + max(LONGITUDE)) / 2)-5 FROM STATION WHERE DISTRICT_ID = '110' AND NAME LIKE 'Vancouver%')
    and
    (SELECT round((min(LONGITUDE) + max(LONGITUDE)) / 2)+5 FROM STATION WHERE DISTRICT_ID = '110' AND NAME LIKE 'Vancouver%')
ORDER BY
  LATITUDE

Question

如何在不使用视图的情况下简化此查询以消除冗余？

限制

数据库是 MySQL，但 ANSI SQL 总是不错的。

谢谢你！

select 
  name
from 
  (select 
    round((min(LATITUDE) + max(LATITUDE)) / 2) as LATITUDE,
    round((min(LONGITUDE) + max(LONGITUDE)) / 2) as LONGITUDE
   from STATION 
   where DISTRICT_ID = '110' 
     AND NAME LIKE 'Vancouver%') AS center
  inner join STATION s
where
  s.DISTRICT_ID = '110' 
  and s.NAME like 'Vancouver%'
  and s.LATITUDE between center.LATITUDE - 5 and center.LATITUDE + 5
  and s.LONGITUDE between center.LONGITUDE - 5 and center.LONGITUDE + 5