Calculate new coordinate x meters and y degree away from one coordinate

Unfortunately, there's no such function provided in the API, so you'll have to write your own.

This site gives several calculations involving latitude/longitude and sample JavaScript code. Specifically, the section titled "Destination point given distance and bearing from start point" shows how to calculate what you're asking.

The JavaScript code is at the bottom of that page and here's one possible way to convert it to Objective-C:

- (double)radiansFromDegrees:(double)degrees{    return degrees * (M_PI/180.0);    }- (double)degreesFromRadians:(double)radians{    return radians * (180.0/M_PI);}- (CLLocationCoordinate2D)coordinateFromCoord:        (CLLocationCoordinate2D)fromCoord         atDistanceKm:(double)distanceKm         atBearingDegrees:(double)bearingDegrees{    double distanceRadians = distanceKm / 6371.0;      //6,371 = Earth's radius in km    double bearingRadians = [self radiansFromDegrees:bearingDegrees];    double fromLatRadians = [self radiansFromDegrees:fromCoord.latitude];    double fromLonRadians = [self radiansFromDegrees:fromCoord.longitude];    double toLatRadians = asin( sin(fromLatRadians) * cos(distanceRadians)         + cos(fromLatRadians) * sin(distanceRadians) * cos(bearingRadians) );    double toLonRadians = fromLonRadians + atan2(sin(bearingRadians)         * sin(distanceRadians) * cos(fromLatRadians), cos(distanceRadians)         - sin(fromLatRadians) * sin(toLatRadians));    // adjust toLonRadians to be in the range -180 to +180...    toLonRadians = fmod((toLonRadians + 3*M_PI), (2*M_PI)) - M_PI;    CLLocationCoordinate2D result;    result.latitude = [self degreesFromRadians:toLatRadians];    result.longitude = [self degreesFromRadians:toLonRadians];    return result;}

In the JS code, it contains this link which shows a more accurate calculation for distances greater than 1/4 of the Earth's circumference.

Also note the above code accepts distance in km so be sure to divide meters by 1000.0 before passing.

I found one way of doing it, had to dig to find the correct structs and functions. I ended up not using degrees but meters for lat and long instead.

Here's how I did it:

-(CLLocationCoordinate2D)translateCoord:(CLLocationCoordinate2D)coord MetersLat:(double)metersLat MetersLong:(double)metersLong{    CLLocationCoordinate2D tempCoord;    MKCoordinateRegion tempRegion = MKCoordinateRegionMakeWithDistance(coord, metersLat, metersLong);    MKCoordinateSpan tempSpan = tempRegion.span;    tempCoord.latitude = coord.latitude + tempSpan.latitudeDelta;    tempCoord.longitude = coord.longitude + tempSpan.longitudeDelta;    return tempCoord;}

And of course, if I really need to use degrees in the future, it's pretty easy (I think...) to do some changes to above to get it to work like I actually asked for.

Using an MKCoordinateRegion has some issues—the returned region can be adjusted to fit since the two deltas may not exactly map to the projection at that latitude, if you want zero delta for one of the axes you are out of luck, etc.

This function uses MKMapPoint to perform coordinate translations which allows you to move points around in the map projection's coordinate space and then extract a coordinate from that.

CLLocationCoordinate2D MKCoordinateOffsetFromCoordinate(CLLocationCoordinate2D coordinate, CLLocationDistance offsetLatMeters, CLLocationDistance offsetLongMeters) {    MKMapPoint offsetPoint = MKMapPointForCoordinate(coordinate);    CLLocationDistance metersPerPoint = MKMetersPerMapPointAtLatitude(coordinate.latitude);    double latPoints = offsetLatMeters / metersPerPoint;    offsetPoint.y += latPoints;    double longPoints = offsetLongMeters / metersPerPoint;    offsetPoint.x += longPoints;    CLLocationCoordinate2D offsetCoordinate = MKCoordinateForMapPoint(offsetPoint);    return offsetCoordinate;}