How do I know if a Lat,Lng point is contained within a circle?

Question

Ok pretty self explanatory. I'm using google maps and I'm trying to find out if a lat,long point is within a circle of radius say x (x is chosen by the user).

Bounding box will not work for this. I have already tried using the following code:

distlatLng = new google.maps.LatLng(dist.latlng[0],dist.latlng[1]);
var latLngBounds = circle.getBounds();
if(latLngBounds.contains(distlatLng)){
      dropPins(distlatLng,dist.f_addr);
}

This still results in markers being places outside the circle.

I'm guess this is some simple maths requiring the calculation of the curvature or an area but I'm not sure where to begin. Any suggestions?

Answer 1

Working solution with dragable center marker

Have you ever tried contains? Take a look at the LatLngBounds Constructor.

I wrote an article about it, that contains a link to a working JSFiddle.net example.

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Updated version.

Answer 2

Unfortunately Pythagoras is no help on a sphere. Thus Stuart Beard's answer is incorrect; longitude differences don't have a fixed ratio to metres but depend on the latitude.

The correct way is to use the formula for great circle distances. A good approximation, assuming a spherical earth, is this (in C++):

/** Find the great-circle distance in metres, assuming a spherical earth, between two lat-long points in degrees. */
inline double GreatCircleDistanceInMeters(double aLong1,double aLat1,double aLong2,double aLat2)
    {
    aLong1 *= KDegreesToRadiansDouble;
    aLat1 *= KDegreesToRadiansDouble;
    aLong2 *= KDegreesToRadiansDouble;
    aLat2 *= KDegreesToRadiansDouble;
    double cos_angle = sin(aLat1) * sin(aLat2) + cos(aLat1) * cos(aLat2) * cos(aLong2 - aLong1);

    /*
    Inaccurate trig functions can cause cos_angle to be a tiny amount
    greater than 1 if the two positions are very close. That in turn causes
    acos to give a domain error and return the special floating point value
    -1.#IND000000000000, meaning 'indefinite'. Observed on VS2008 on 64-bit Windows.
    */
    if (cos_angle >= 1)
        return 0;

    double angle = acos(cos_angle);
    return angle * KEquatorialRadiusInMetres;
    }

where

const double KPiDouble = 3.141592654;
const double KDegreesToRadiansDouble = KPiDouble / 180.0;

and

/**
A constant to convert radians to metres for the Mercator and other projections.
It is the semi-major axis (equatorial radius) used by the WGS 84 datum (see http://en.wikipedia.org/wiki/WGS84).
*/
const int32 KEquatorialRadiusInMetres = 6378137;