transfom k m into gps Coordonates

Answer 1

Lat/Lng coordinates don't work with the Phytagoras Formula, as Earth is a sphere, and not flat...

Take a look at this manual:

http://www.movable-type.co.uk/scripts/latlong.html

You need the distance equation, to see if the distance between the lat/lng you have and the lat/lng of the center of the circle is smaller than R.

Answer 2

It's not that simple. You can't generally use the euclidean distance. This is how i think you could do it:

The correct way

The correct way to do this would be to calculate the great circle distance (the shortest walking distance on the surface), which I never really understood, but it's pretty easy to google (and adamk already gave you a link to it).

The lazy way

Assume that your points are not too close to the poles and that they are close to each other. Then you can use the latitude and longitude as if they were euclidean coordinates. (this is basically some cylindrical projection). 1 degree of latitude will then be (earth_circumference / 360) long and 1 degree of longitude will be (cos(lat) * earth_circumference / 360) long.

the complete code would look something like this:

double distance_lazy(double lat1, double lon1, double lat2, double lon2){
    double xDist = (lat2 - lat1) * EARTH_CIRCUMFERENCE / 360.0;
    double yDist = cos(lat1) * (lon2 - lon1) * EARTH_CIRCUMFERENCE / 360.0;
    return sqrt(xDist^2 + yDist^2);
}

If your distances will be only a few km from each other, this should work pretty ok for the whole europe ... something like that.

The "weird definition" way

Another thing you can say is, that "distance" is the length straight line between the points, even though it goes through the earth. Then this would become calculating the 3D coordinates of a point and then their euclidean distance

double distance_straight_line(double lat1, double lon1, double lat2, double lon2){
    double x1 = cos(lat1) * cos(lon1) * EARTH_RADIUS;
    double y1 = sin(lat1) * cos(lon1) * EARTH_RADIUS;
    double z1 = sin(lon1) * EARTH_RADIUS;

    double x2 = cos(lat2) * cos(lon2) * EARTH_RADIUS;
    double y2 = sin(lat2) * cos(lon2) * EARTH_RADIUS;
    double z2 = sin(lon2) * EARTH_RADIUS;

    return sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2);
}

Again this will work as expected for points close to each other, this time they can be anywhere in the world. If you give it points far from each other, the output will be correct, but prety useless (unless you are very good at digging).

Hope it helps.