>>This is likely related to your latitude, longitude question in SQL server. If so, again I would use MSSQL's spatial data methods. ie:
>>
>>

DECLARE @g geography;
>>SET @g1 = geography::STGeomFromText('POINT(-65.7142 47.7661)', 4326); --Petit-Rocher nord
>>SET @g2 = geography::STGeomFromText('POINT(-65.7305 47.7975)', 4326); --Petit-Rocher sud
>>
>>select @g1.STDistance(@g2);
>>
>>SELECT @g1.STBuffer(3700).STIntersects(@g2); -- true, within 3700 meters

Cetin
>
>Given this:
>
>

>DECLARE @CenterLat Float=47.7795
>DECLARE @CenterLon Float=-65.7191
>DECLARE @SearchDistance Float=250
>
>SELECT Result.Latitude,Result.Longitude
> FROM (SELECT * FROM Test T
> CROSS APPLY (SELECT SIN(@CenterLat/57.2957795130823)*SIN(Latitude/57.2957795130823)+
> COS(@CenterLat/57.2957795130823)*COS(Latitude/57.2957795130823)*
> COS(Longitude/57.2957795130823-@CenterLon/57.2957795130823) AS Temp) Temp
> WHERE (Latitude>=@CenterLat-@SearchDistance/111.0 AND Latitude<=@CenterLat+@SearchDistance/111.0) AND
> (Longitude>=@CenterLon-@SearchDistance/111.0 AND Longitude<=@CenterLon+@SearchDistance/111.0)) Result
> WHERE 3958.75586574*ACOS(CASE WHEN Temp > 1 THEN 1 WHEN Temp < -1 THEN -1 ELSE Temp END)<=@SearchDistance
>

>
>How can we adjust that SQL to benefit of this geography approach?

DECLARE @g geography,
    @area geography ;
SET @g = geography::Point(47.7795,-65.7191,4326) ; --Petit-Rocher
SET @area = @g.STBuffer(250) ;

SELECT  locationId,
        Latitude,
        Longitude,
        @g.STDistance(geoPoint) AS distance
FROM    ( SELECT    locationId,
                    Latitude,
                    Longitude,
                    geography::Point(Latitude,Longitude,4326) AS geoPoint
          FROM      result
        ) geoData
WHERE   @area.STIntersects(geoData.geoPoint) = 1 ;

PS: If you need more speed then consider storing as spatial data and indexing spatial data as explained in SQL 2008 documentation.
Cetin