Get all points coordinate on a MKCircle

Question

I am drawing a MKCircle on MKMapView, I need to get all points in the Circle border (not inside it) to send the points coordinates to server and get the information that related to the region which is inside the circle.

Also as I have the center coordinate and radius of circle So maybe points could be calculated by this variables.

How can I get the points coordinates ?

Answer 1

I calculated the points, But as you know we can't get ALL the points, But this method returns about 40 points of the circle that is enough.

internal func radiusSearchPoints(coordinate: CLLocationCoordinate2D, radius: CLLocationDistance) -> [CLLocationCoordinate2D] {
    var points: [CLLocationCoordinate2D] = []
    let earthRadius = 6_378_100.0
    let π = Double.pi
    let lat = coordinate.latitude * π / 180.0
    let lng = coordinate.longitude * π / 180.0

    var t: Double = 0
    while t <= 2 * π {
        let pointLat = lat + (radius / earthRadius) * sin(t)
        let pointLng = lng + (radius / earthRadius) * cos(t)

        let point = CLLocationCoordinate2D(latitude: pointLat * 180 / π, longitude: pointLng * 180 / π)
        points.append(point)
        t += 0.3
    }

    return points
}

Answer 2

There are infinite points on the circumference of a circle, so it's quite impossible to get all of them.

However, it is possible to determine if a given point is on the circle's border using just your centre coordinate and your radius

The equation of a circle at the origin is given by the parametric equation:

x = cos(angle) * radius
y = sin(angle) * radius

For the example circle: centre = (1, 1) radius = 2 and the example point (3, 1)

dx = point.x - centre.x
dy = point.y - centre.y

dx = 3 - 1 = 2
dy = 1 - 1 - 0

angle = atan2(dy/dx) = 0

dx = cos(angle) * radius
rearranging gives:

radius = dx/cos(angle)
radius = 2/cos(0)
radius = 2/1
radius = 2 

The distance between this point and the centre of the circle is the same as the radius, therefore the point is on the circumference.