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Find intersection coordinates of two circles on earth?

I'm trying to find a second intersection point of two circles. One of the points that I already know was used to calculate a distance and then used as the circle radius (exemple). The problem is that im not getting the know point, im getting two new coordinates, even thou they are similar. The problem is probably related to the earth curvature but I have searched for some solution and found nothing.
The circles radius are calculated with the earth curvature. And this is the code I have:
function GET_coordinates_of_circles(position1,r1, position2,r2) {
var deg2rad = function (deg) { return deg * (Math.PI / 180); };
x1=position1.lng;
y1=position1.lat;
x2=position2.lng;
y2=position2.lat;
var centerdx = deg2rad(x1 - x2);
var centerdy = deg2rad(y1 - y2);
var R = Math.sqrt(centerdx * centerdx + centerdy * centerdy);
if (!(Math.abs(r1 - r2) <= R && R <= r1 + r2)) { // no intersection
console.log("nope");
return []; // empty list of results
}
// intersection(s) should exist
var R2 = R*R;
var R4 = R2*R2;
var a = (r1*r1 - r2*r2) / (2 * R2);
var r2r2 = (r1*r1 - r2*r2);
var c = Math.sqrt(2 * (r1*r1 + r2*r2) / R2 - (r2r2 * r2r2) / R4 - 1);
var fx = (x1+x2) / 2 + a * (x2 - x1);
var gx = c * (y2 - y1) / 2;
var ix1 = fx + gx;
var ix2 = fx - gx;
var fy = (y1+y2) / 2 + a * (y2 - y1);
var gy = c * (x1 - x2) / 2;
var iy1 = fy + gy;
var iy2 = fy - gy;
// note if gy == 0 and gx == 0 then the circles are tangent and there is only one solution
// but that one solution will just be duplicated as the code is currently written
return [[iy1, ix1], [iy2, ix2]];
}
The deg2rad variable it is suppose to adjust the other calculations with the earth curvature.
Thank you for any help.

Your calculations for R and so on are wrong because plane Pythagorean formula does not work for spherical trigonometry (for example - we can have triangle with all three right angles on the sphere!). Instead we should use special formulas. Some of them are taken from this page.
At first find big circle arcs in radians for both radii using R = Earth radius = 6,371km
a1 = r1 / R
a2 = r2 / R
And distance (again arc in radians) between circle center using haversine formula
var R = 6371e3; // metres
var φ1 = lat1.toRadians();
var φ2 = lat2.toRadians();
var Δφ = (lat2-lat1).toRadians();
var Δλ = (lon2-lon1).toRadians();
var a = Math.sin(Δφ/2) * Math.sin(Δφ/2) +
Math.cos(φ1) * Math.cos(φ2) *
Math.sin(Δλ/2) * Math.sin(Δλ/2);
var ad = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
And bearing from position1 to position 2:
//where φ1,λ1 is the start point, φ2,λ2 the end point
//(Δλ is the difference in longitude)
var y = Math.sin(λ2-λ1) * Math.cos(φ2);
var x = Math.cos(φ1)*Math.sin(φ2) -
Math.sin(φ1)*Math.cos(φ2)*Math.cos(λ2-λ1);
var brng = Math.atan2(y, x);
Now look at the picture from my answer considering equal radii case.
(Here circle radii might be distinct and we should use another approach to find needed arcs)
We have spherical right-angle triangles ACB and FCB (similar to plane case BD is perpendicular to AF in point C and BCA angle is right).
Spherical Pythagorean theorem (from the book on sph. trig) says that
cos(AB) = cos(BC) * cos(AC)
cos(FB) = cos(BC) * cos(FC)
or (using x for AC, y for BC and (ad-x) for FC)
cos(a1) = cos(y) * cos(x)
cos(a2) = cos(y) * cos(ad-x)
divide equations to eliminate cos(y)
cos(a1)*cos(ad-x) = cos(a2) * cos(x)
cos(a1)*(cos(ad)*cos(x) + sin(ad)*sin(x)) = cos(a2) * cos(x)
cos(ad)*cos(x) + sin(ad)*sin(x) = cos(a2) * cos(x) / cos(a1)
sin(ad)*sin(x) = cos(a2) * cos(x) / cos(a1) - cos(ad)*cos(x)
sin(ad)*sin(x) = cos(x) * (cos(a2) / cos(a1) - cos(ad))
TAC = tg(x) = (cos(a2) / cos(a1) - cos(ad)) / sin(ad)
Having hypotenuse and cathetus of ACB triangle we can find angle between AC and AB directions (Napier's rules for right spherical triangles) - note we already know TAC = tg(AC) and a1 = AB
cos(CAB)= tg(AC) * ctg(AB)
CAB = Math.acos(TAC * ctg(a1))
Now we can calculate intersection points - they lie at arc distance a1 from position1 along bearings brng-CAB and brng+CAB
B_bearing = brng - CAB
D_bearing = brng + CAB
Intersection points' coordinates:
var latB = Math.asin( Math.sin(lat1)*Math.cos(a1) +
Math.cos(lat1)*Math.sin(a1)*Math.cos(B_bearing) );
var lonB = lon1.toRad() + Math.atan2(Math.sin(B_bearing)*Math.sin(a1)*Math.cos(lat1),
Math.cos(a1)-Math.sin(lat1)*Math.sin(lat2));
and the same for D_bearing
latB, lonB are in radians

I had a similar need ( Intersection coordinates (lat/lon) of two circles (given the coordinates of the center and the radius) on earth ) and hereby I share the solution in python in case it might help someone:
'''
FINDING THE INTERSECTION COORDINATES (LAT/LON) OF TWO CIRCLES (GIVEN THE COORDINATES OF THE CENTER AND THE RADII)
Many thanks to Ture Pålsson who directed me to the right source, the code below is based on whuber's brilliant work here:
https://gis.stackexchange.com/questions/48937/calculating-intersection-of-two-circles
The idea is that;
1. The points in question are the mutual intersections of three spheres: a sphere centered beneath location x1 (on the
earth's surface) of a given radius, a sphere centered beneath location x2 (on the earth's surface) of a given radius, and
the earth itself, which is a sphere centered at O = (0,0,0) of a given radius.
2. The intersection of each of the first two spheres with the earth's surface is a circle, which defines two planes.
The mutual intersections of all three spheres therefore lies on the intersection of those two planes: a line.
Consequently, the problem is reduced to intersecting a line with a sphere.
Note that "Decimal" is used to have higher precision which is important if the distance between two points are a few
meters.
'''
from decimal import Decimal
from math import cos, sin, sqrt
import math
import numpy as np
def intersection(p1, r1_meter, p2, r2_meter):
# p1 = Coordinates of Point 1: latitude, longitude. This serves as the center of circle 1. Ex: (36.110174, -90.953524)
# r1_meter = Radius of circle 1 in meters
# p2 = Coordinates of Point 2: latitude, longitude. This serves as the center of circle 1. Ex: (36.110174, -90.953524)
# r2_meter = Radius of circle 2 in meters
'''
1. Convert (lat, lon) to (x,y,z) geocentric coordinates.
As usual, because we may choose units of measurement in which the earth has a unit radius
'''
x_p1 = Decimal(cos(math.radians(p1[1]))*cos(math.radians(p1[0]))) # x = cos(lon)*cos(lat)
y_p1 = Decimal(sin(math.radians(p1[1]))*cos(math.radians(p1[0]))) # y = sin(lon)*cos(lat)
z_p1 = Decimal(sin(math.radians(p1[0]))) # z = sin(lat)
x1 = (x_p1, y_p1, z_p1)
x_p2 = Decimal(cos(math.radians(p2[1]))*cos(math.radians(p2[0]))) # x = cos(lon)*cos(lat)
y_p2 = Decimal(sin(math.radians(p2[1]))*cos(math.radians(p2[0]))) # y = sin(lon)*cos(lat)
z_p2 = Decimal(sin(math.radians(p2[0]))) # z = sin(lat)
x2 = (x_p2, y_p2, z_p2)
'''
2. Convert the radii r1 and r2 (which are measured along the sphere) to angles along the sphere.
By definition, one nautical mile (NM) is 1/60 degree of arc (which is pi/180 * 1/60 = 0.0002908888 radians).
'''
r1 = Decimal(math.radians((r1_meter/1852) / 60)) # r1_meter/1852 converts meter to Nautical mile.
r2 = Decimal(math.radians((r2_meter/1852) / 60))
'''
3. The geodesic circle of radius r1 around x1 is the intersection of the earth's surface with an Euclidean sphere
of radius sin(r1) centered at cos(r1)*x1.
4. The plane determined by the intersection of the sphere of radius sin(r1) around cos(r1)*x1 and the earth's surface
is perpendicular to x1 and passes through the point cos(r1)x1, whence its equation is x.x1 = cos(r1)
(the "." represents the usual dot product); likewise for the other plane. There will be a unique point x0 on the
intersection of those two planes that is a linear combination of x1 and x2. Writing x0 = ax1 + b*x2 the two planar
equations are;
cos(r1) = x.x1 = (a*x1 + b*x2).x1 = a + b*(x2.x1)
cos(r2) = x.x2 = (a*x1 + b*x2).x2 = a*(x1.x2) + b
Using the fact that x2.x1 = x1.x2, which I shall write as q, the solution (if it exists) is given by
a = (cos(r1) - cos(r2)*q) / (1 - q^2),
b = (cos(r2) - cos(r1)*q) / (1 - q^2).
'''
q = Decimal(np.dot(x1, x2))
if q**2 != 1 :
a = (Decimal(cos(r1)) - Decimal(cos(r2))*q) / (1 - q**2)
b = (Decimal(cos(r2)) - Decimal(cos(r1))*q) / (1 - q**2)
'''
5. Now all other points on the line of intersection of the two planes differ from x0 by some multiple of a vector
n which is mutually perpendicular to both planes. The cross product n = x1~Cross~x2 does the job provided n is
nonzero: once again, this means that x1 and x2 are neither coincident nor diametrically opposite. (We need to
take care to compute the cross product with high precision, because it involves subtractions with a lot of
cancellation when x1 and x2 are close to each other.)
'''
n = np.cross(x1, x2)
'''
6. Therefore, we seek up to two points of the form x0 + t*n which lie on the earth's surface: that is, their length
equals 1. Equivalently, their squared length is 1:
1 = squared length = (x0 + t*n).(x0 + t*n) = x0.x0 + 2t*x0.n + t^2*n.n = x0.x0 + t^2*n.n
'''
x0_1 = [a*f for f in x1]
x0_2 = [b*f for f in x2]
x0 = [sum(f) for f in zip(x0_1, x0_2)]
'''
The term with x0.n disappears because x0 (being a linear combination of x1 and x2) is perpendicular to n.
The two solutions easily are t = sqrt((1 - x0.x0)/n.n) and its negative. Once again high precision
is called for, because when x1 and x2 are close, x0.x0 is very close to 1, leading to some loss of
floating point precision.
'''
if (np.dot(x0, x0) <= 1) & (np.dot(n,n) != 0): # This is to secure that (1 - np.dot(x0, x0)) / np.dot(n,n) > 0
t = Decimal(sqrt((1 - np.dot(x0, x0)) / np.dot(n,n)))
t1 = t
t2 = -t
i1 = x0 + t1*n
i2 = x0 + t2*n
'''
7. Finally, we may convert these solutions back to (lat, lon) by converting geocentric (x,y,z) to geographic
coordinates. For the longitude, use the generalized arctangent returning values in the range -180 to 180
degrees (in computing applications, this function takes both x and y as arguments rather than just the
ratio y/x; it is sometimes called "ATan2").
'''
i1_lat = math.degrees( math.asin(i1[2]))
i1_lon = math.degrees( math.atan2(i1[1], i1[0] ) )
ip1 = (i1_lat, i1_lon)
i2_lat = math.degrees( math.asin(i2[2]))
i2_lon = math.degrees( math.atan2(i2[1], i2[0] ) )
ip2 = (i2_lat, i2_lon)
return [ip1, ip2]
elif (np.dot(n,n) == 0):
return("The centers of the circles can be neither the same point nor antipodal points.")
else:
return("The circles do not intersect")
else:
return("The centers of the circles can be neither the same point nor antipodal points.")
'''
Example: The output of below is [(36.989311051533505, -88.15142628069133), (38.2383796094578, -92.39048549120287)]
intersection_points = intersection((37.673442, -90.234036), 107.5*1852, (36.109997, -90.953669), 145*1852)
print(intersection_points)
'''
Any feedback is appreciated.

Which one of these azimuth values is the right one? And why?

First of all, let's make clear that I want the Azimuth on the surface of the Earth, i.e. the angle between two locations, for example New York and Moscow.
I am testing some azimuth calculations with my JS functions (shown below). For the points A(-170, -89) to B(10, 89), I get ~90º.
JS function for Azimuth on sphere (from Wikipedia)
var dLon = lon2 - lon1;
var y = Math.sin(dLon) * Math.cos(lat2);
var x = Math.cos(lat1) * Math.sin(lat2) - Math.sin(lat1) * Math.cos(lat2) * Math.cos(dLon);
var angle = Math.atan2(y, x) * 180 / Math.PI;
JS function for Azimuth on oblate spheroid (from Wikipedia)
var dLon = lon2 - lon1;
var f = 1 / 298.257223563; /* Flattening for WGS 84 */
var b = (1 - f) * (1 - f);
var tanLat2 = Math.tan(lat2);
var y = Math.sin(dLon);
var x;
if (lat1 === 0) {
var x = b * tanLat2;
} else {
var a = f * (2 - f);
var tanLat1 = Math.tan(lat1);
var c = 1 + b * tanLat2 * tanLat2;
var d = 1 + b * tanLat1 * tanLat1;
var t = b * Math.tan(lat2) / Math.tan(lat1) + a * Math.sqrt(c / d);
var x = (t - Math.cos(dLon)) * Math.sin(lat1);
}
var angle = Math.atan2(y, x) * 180 / Math.PI;
In Calculator 2, I get 90º.
In PostGIS, I get 270º
In Calculator 1, I get 180º.
I know the Azimuth gets more and more distorted near the Poles, but that's exactly why I am testing at these spots. This variety of different solutions are confusing me. Could you please help me getting the right answer for this?

It depends on the reference used for azimuth, e.g. map-types use 0° for North and positive is clockwise, while math-types uses 0° for East and positive is anticlockwise.
The pair of coordinates A(-170, -89) and B(10, 89) are antipodes which are a special case for finding minimum distances and azimuths. Your question can be answered with a thought exercise.
First note that the half-circumferences of the earth are:
Equatorial: 20037.5085 km
Meridional (north-to-south): 20003.93 km
For a pair of antipodes on the north and south poles, there are an infinite number of azimuths, since the distance is the same along each longitude. (What direction would you go from the south pole to the north pole?)
For a pair of antipodes on the equator, the shortest distances are north or south, since it is slightly shorter along the meridional direction.
For any other pair of antipodes, it is the same answer as from the equator: north or south.
Update
To investigate the problem a bit more with the PostGIS SQL query:
SELECT ST_Distance(A, B), degrees(ST_Azimuth(A, B))
FROM (
SELECT 'POINT(-170 -89)'::geography A, 'POINT(10 89)'::geography B
) f;
With PostGIS 2.0 and 2.1, the incorrect results are:
st_distance | degrees
-----------------+------------------
20003900.583699 | 270.005278779849
But with PostGIS 2.2 (and PROJ 4.9.1), the correct results are now:
st_distance | degrees
------------------+---------
20003931.4586255 | 180

JavaScript function nearest geographical neighbor

I am looking for a JavaScript function who returns the nearest neighbor of a number. e.g: I am having a coordinate 12,323432/12,234223 and i want to know the nearest coordinate of a set of 20 other coordinates in a database.
How to handle that?

The following 3 functions find the nearest coordinate from a javascript array using the Haversine formula.
function toRad(Value) {
/** Converts numeric degrees to radians */
return Value * Math.PI / 180;
}
function haversine(lat1,lat2,lng1,lng2){
rad = 6372.8; // for km Use 3961 for miles
deltaLat = toRad(lat2-lat1);
deltaLng = toRad(lng2-lng1);
lat1 = toRad(lat1);
lat2 = toRad(lat2);
a = Math.sin(deltaLat/2) * Math.sin(deltaLat/2) + Math.sin(deltaLng/2) * Math.sin(deltaLng/2) * Math.cos(lat1) * Math.cos(lat2);
c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
return rad * c;
}
function calculate(){
var result = haversine(lat1,coordArray [0][0],lng1,coordArray [0][1]);
for (var i=1;i<coordArray.length;i++){
var ans = haversine(lat1,coordArray [i][0],lng1,coordArray [i][1]);
if (ans < result){//nearest
result = ans;
}
}
document.write("Result " +result);
}

Less rounding errors with the Vincenty formula
The Haversine formula suffers from rounding errors for the special (and somewhat unusual) case of antipodal points (on opposite ends of the sphere).
A better choice is therefore the Vincenty formula for the special case of a sphere. It is computationally not more demanding, but suffers less from machine rounding errors.
Here is my Python3 implementation, which runs in any browser using Brython or which one can easily manually transcode to JavaScript:
from math import radians, sin, cos, atan2, sqrt
def vincenty_sphere(lat1,lat2,lon1,lon2):
lat1 = radians(lat1)
lat2 = radians(lat2)
delta_lon = radians(lon2-lon1)
term1 = (cos(lat2) * sin(delta_lon))**2
term2 = (cos(lat1) * sin(lat2) - sin(lat1) * cos(lat2) * cos(delta_lon))**2
numerator = sqrt(term1 + term2)
denominator = sin(lat1) * sin(lat2) + cos(lat1) * cos(lat2) * cos(delta_lon)
central_angle = atan2(numerator, denominator)
radius = 6372.8 # km
return radius * central_angle
def station_near(geo):
lat = geo['latitude']
lon = geo['longitude']
nearest = 40042.0 # km
for s in range(len(STATIONS)):
distance = vincenty_sphere(lat, STATIONS[s].lat, lon, STATIONS[s].lon)
if(distance < nearest):
nearest = distance
station = s
return station

How to find my distance to a known location in JavaScript

Using JavaScript in the browser, how can I determine the distance from my current location to another location for which I have the latitude and longitude?

If your code runs in a browser, you can use the HTML5 geolocation API:
window.navigator.geolocation.getCurrentPosition(function(pos) {
console.log(pos);
var lat = pos.coords.latitude;
var lon = pos.coords.longitude;
})
Once you know the current position and the position of your "target", you can calculate the distance between them in the way documented in this question: Calculate distance between two latitude-longitude points? (Haversine formula).
So the complete script becomes:
function distance(lon1, lat1, lon2, lat2) {
var R = 6371; // Radius of the earth in km
var dLat = (lat2-lat1).toRad(); // Javascript functions in radians
var dLon = (lon2-lon1).toRad();
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(lat1.toRad()) * Math.cos(lat2.toRad()) *
Math.sin(dLon/2) * Math.sin(dLon/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c; // Distance in km
return d;
}
/** Converts numeric degrees to radians */
if (typeof(Number.prototype.toRad) === "undefined") {
Number.prototype.toRad = function() {
return this * Math.PI / 180;
}
}
window.navigator.geolocation.getCurrentPosition(function(pos) {
console.log(pos);
console.log(
distance(pos.coords.longitude, pos.coords.latitude, 42.37, 71.03)
);
});
Apparently I am 6643 meters from the center of Boston, MA right now (that's the hard-coded second location).
See these links for more information:
http://html5demos.com/geo
http://diveintohtml5.info/geolocation.html
Calculate distance between two latitude-longitude points? (Haversine formula)
toRad() Javascript function throwing error

Calculate distance for multiple locations google map

Assume I have 3 locations namely Kallang(A), Paya Lebar(B) and Bedok(C). I try to do a A->B + B->C but the distance always return me A->C. Why is this so ??
kallang to paya lebar(A-B) -> 2.5914199062350742 km
paya lebar to bedok(B-C) -> 4.4403012109180775 km
total (A-B-C) -> 7.03 km
kallang to bedok (A-C) -> 6.88 km
below are my codes:
var R = 6371;
var dLat = toRad(location3.lat()-location2.lat()) + toRad(location2.lat()-location1.lat());
var dLon = toRad(location3.lng()-location2.lng()) + toRad(location2.lng()-location1.lng());
var dLat1 = toRad(location1.lat());
var dLat2 = toRad(location2.lat());
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(dLat1) * Math.cos(dLat1) *
Math.sin(dLon/2) * Math.sin(dLon/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
document.getElementById("distance_direct").innerHTML = "<br/><br/><br/>The distance between the five points (in a straight line) is: "+d +" km.";
}
function toRad(deg)
{
return deg * Math.PI/180;
}
Is it something wrong with my adding and subtracting calculation statement ??? Thanks

You can just use the built in computeDistanceBetween method from the geometry library

Yep, the problem is with adding and substracting the same thing. Since toRad() is just a multiplication, it's associative respective to + and - and therefore
var dLat = toRad(location3.lat()-location2.lat())
+ toRad(location2.lat()-location1.lat());
is exactly the same as
var dLat = toRad(location3.lat()-location1.lat());
so you end up calculating the straight distance between the first and last point.
If you are using the Google Maps JavaScript API V3, you can just load the Geometry Library and do this:
var indirect_distance =
google.maps.geometry.spherical.computeDistanceBetween(location1, location2)
+ google.maps.geometry.spherical.computeDistanceBetween(location2, location3);