Find if a point lies inside a Circle PHP

Question

If I have coordinates of center point of circle, circle radius and also coordinates of the point that need to be check if a point lies inside or on the boundary of the Circle.

$circle_x and $circle_y => center point latitude and longitude of the circle
$rad => radius(Meter) of the circle
$x and $y => latitude and longitude of the point that need to be check

I have tried 2 ways,

    public function pointInCircle($circle_x, $circle_y, $rad, $x, $y)
    {
        $dx = abs($x - $circle_x);
        $dy = abs($y - $circle_y);
        $R = $rad;

        if ($dx + $dy <= $R) {
            return "inside";
        } else if ($dx > $R) {
            return "outside";
        } else if ($dy > $R) {
            return "outside";
        } else if ($dx ^ 2 + $dy ^ 2 <= $R ^ 2) {
            return "inside";
        } else {
            return "outside";
        }
    }

Second method is,

    public function pointInCircle($circle_x, $circle_y, $rad, $x, $y)
    {
        if ((($x - $circle_x) * ($x - $circle_x) + ($y - $circle_y) * ($y - $circle_y)) <= ($rad * $rad)){
            return "inside";
        } else {
            return "outside";
        }
    }

But above 2 methods couldn't provide correct result, please help to find if provided point lies inside or on the boundary of the Circle using PHP.

PHP code is all correct but equation use may not be, (circleX,circleY) point of center of circle & (x,y) point is in circle or out? if (d^2 = (x-circleX)^2 + (y-circleY)^2) >= radius then point in circle else not in circle. I think is the only check need. — MZM, Nov 30 '22 at 04:53
what is the 'd' that you mentioned and can we pass radius using meters — Shasika Madhushan, Nov 30 '22 at 04:56
if (((($x-$centerX) * ($x-$centerX)) + (($y-$centerY) * ($y-$centerY))) >= $radius) { return true; } — Shasika Madhushan, Nov 30 '22 at 05:21

MBo · Accepted Answer · 2022-11-30T05:54:48.720

2

To solve this problem on the earth surface with latidute/longitude coordinates, you need to apply spherical geometry.

Look here at Distance section and modify code for PHP.

P.S. Found PHP implementation here at SO.

If result distance is less than your $rad, point with given coordinates lies inside the circle

function haversineGreatCircleDistance(
  $latitudeFrom, $longitudeFrom, $latitudeTo, $longitudeTo, $earthRadius = 6371000)
{
  // convert from degrees to radians
  $latFrom = deg2rad($latitudeFrom);
  $lonFrom = deg2rad($longitudeFrom);
  $latTo = deg2rad($latitudeTo);
  $lonTo = deg2rad($longitudeTo);

  $latDelta = $latTo - $latFrom;
  $lonDelta = $lonTo - $lonFrom;

  $angle = 2 * asin(sqrt(pow(sin($latDelta / 2), 2) +
    cos($latFrom) * cos($latTo) * pow(sin($lonDelta / 2), 2)));
  return $angle * $earthRadius;
}

now

if (haversineGreatCircleDistance($circle_y, $circle_x, $y, $x) < $rad)
    point i circle

Note - I supposed that latitude corresponds to Y-coordinate, and longitude - to X-coordinates, as maps usually are drawn. Swap x/y if I'm not right

(Test here) gives OUTSIDE for small $rad and Inside for large radius

$circle_y = 48.8588336; //Paris center
$circle_x = 2.2769953;

$y = 48.80;   //Versailles
$x = 2.084;
//$rad = 1000;
$rad = 100000;

if (haversineGreatCircleDistance($circle_y, $circle_x, $y, $x) < $rad)
   echo "Inside";
else
   echo "OUTSIDE";

edited Nov 30 '22 at 05:54

answered Nov 30 '22 at 05:21

MBo

77,366
5
53
86

but i need this in PHP side can we do that? – Shasika Madhushan Nov 30 '22 at 05:23
This method also provide always true (inside) – Shasika Madhushan Nov 30 '22 at 05:43
@Shasika Madhushan could you show data used for checking? – MBo Nov 30 '22 at 05:44
$circle_y -> 32.809136, $circle_x -> 150.325108, $rad -> 5000.0, $y -> -32.8743, $x -> 149.8236 – Shasika Madhushan Nov 30 '22 at 05:47
according to above data I need false result(outside) – Shasika Madhushan Nov 30 '22 at 05:54
Yes, code at my link gives outside – MBo Nov 30 '22 at 05:57
Yeah, my mistake I had use $earthRadius as my circle radius – Shasika Madhushan Nov 30 '22 at 06:00
thankyou MBo, it work like carm – Alauddin Afif Cassandra Apr 17 '23 at 06:59

score 0 · Answer 2 · answered Nov 30 '22 at 05:25

0

Broke down the code (partially the code you provided) into separate line to easily debug as needed.

function pointInCircle($circle_x, $circle_y, $rad, $x, $y)
    {
        $dx = ($x - $circle_x); 
        $dy = ($y - $circle_y);
        $dx = $dx * $dx;
        $dy = $dy * $dy;
        
        $dist = $dx + $dy;  //eqt of distance.

       
        $str =  "outside";
        if ( $dist <= $rad * $rad)
            $str = "inside";
        
        return $str;
      
    }
    
    echo " result " . pointInCircle(0,0,25,4,4);

answered Nov 30 '22 at 05:25

MZM

105
1
3

The main problem - author uses latitude and longitude as cartesian coordinates – MBo Nov 30 '22 at 05:27
This also provide always "inside as a return – Shasika Madhushan Nov 30 '22 at 05:29

Find if a point lies inside a Circle PHP

2 Answers2