Point within and on circle

Question

I have solved this problem but im not sure if its correct..

User should give the coordinates of a point and I should check if that point is within,outside or on the circle. I used the distance formula to solve this . The given information about the circle are:

Circle is centered at ( 0,0 ) and radius is 10

 public static void main(String[] strings) {


    Scanner scanner = new Scanner(System.in);

    System.out.println("Enter a point with two coordinates");
    double y1 = scanner.nextDouble();
    double x1 = scanner.nextDouble();

    //  circle is centered at 0,0
    double y2 = 0;
    double x2 = 0;

    double radius = 10;

    double distance;
    // using the distance formula to calculate the distance
    // from the point given from the user and point where circle is centered

    /**
     * distance formula
     *  d = √ ( x2 - x1 )^2 + (y2 - y1 )^2
     */

    distance = Math.pow( x2 - x1,2) + Math.pow(y2-y1,2);

    // find square root
    distance = Math.sqrt(distance);

    String result="";

    if(distance < radius) {
        result = "("+y1+" , "+x1+")" +" is within the circle";
    }
    else if(distance > radius) {
        result = y1+" , "+x1 +" is outside the circle";
    }
    else if(distance == radius) {
        result =y1+" , "+x1 +" is on the circle";
    }

    System.out.println(result);

}

Well, is it correct? Does it give the expected output for the inputs you have tried? — Andy Turner, Mar 10 '16 at 10:01
Whats the question? Where's the error? Looks fine to me so far — f1sh, Mar 10 '16 at 10:03
The logic part looks correct to me and I do not see any errors with your code. Have you tried running it? — user2004685, Mar 10 '16 at 10:04
Checking equality of doubles and floats is not safe: http://stackoverflow.com/questions/7180952/is-checking-a-double-for-equality-ever-safe — MattSmith, Mar 10 '16 at 10:06
For input 4,5 and 9,9 the output is the expected but I dont have a way to check the last case in which a point is on the circle . — Evi Tinou, Mar 10 '16 at 10:06
@EviTinou "For input 4,5 and 9,9 the output ..." you should update the question to say that. — Austin, Mar 10 '16 at 10:18
@EviTinou try a point like (10, 0) - that should be exactly on the circle. — Andy Turner, Mar 10 '16 at 10:22
Are you sure this question requires doubles as input? The examples are integers. — weston, Mar 10 '16 at 11:01

score 3 · Accepted Answer · answered Mar 10 '16 at 10:07

It's fine but sloppy.

There's no need to compute the square root. Work in units of distance squared.

Then compare using distance < radius * radius etc., perhaps renaming distance for the sake of clarity. Computing square roots is costly and imprecision can creep in which can be difficult to control. This is particularly important in your case where you want to test for a point being on the circle's edge.

Also consider writing (x2 - x1) * (x2 - x1) rather than using pow for the second power. Although Java possibly (I never remember for sure which is certainly a good enough reason for my not using it) optimises to the longer form, other languages (such as C) don't and imprecision can creep in there too.

For a beginner, I think that readability is far more important than performance and accuracy concerns. — Andy Turner, Mar 10 '16 at 10:11
Sadly this is not a beginner's problem. It has some tricky floating point considerations. — Bathsheba, Mar 10 '16 at 10:11

Alex Salauyou · Answer 2 · 2016-03-11T10:30:22.210

0

Consider using Math.hypot() to calculate a distance and compare double values using some small threshold:

static final double DELTA = 1E-5;  // not necessarily 1E-5; see comments

//...

distance = Math.hypot(x2 - x1, y2 - y1);    
if (Math.abs(distance - radius) < DELTA)
   // on the circle
else if (distance > radius)
   // outside
else
   // inside

The reason for using DELTA is that there is a very small chance to get equal double values by calculations. If they differ in at least one bit, direct comparison will return false.

By appliying threshold, you check not whether the point lays exactly on the circle, but whether it lays inside a ring between radius - DELTA and radius + DELTA. So DELTA is a kind of tolerance limit which value should be chosen particularily for application, e. g. depending on absolute or relative input inaccuracy.

edited Mar 11 '16 at 10:30

answered Mar 10 '16 at 10:11

Alex Salauyou

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And the *science* behind that choice of `DELTA` is? – Bathsheba Mar 10 '16 at 10:11
@Bathsheba sorry, I'm not going to fall into long discussion. You downvoted thus expressing your attitude to my answer and explanation, and I accepted it. – Alex Salauyou Mar 10 '16 at 10:24
Um. I didn't downvote; not yet anyway. I'm merely inviting you to discuss your choice of `DELTA`. For a small circle, that's a very thick edge ;-) – Bathsheba Mar 10 '16 at 10:27
@Bathsheba okay, sorry for that. No particular reason, just some number that covers input inaccuracy (I believe a user won't enter numbers with more than 5 decimal digits, anyway, it should be selected and explained in every particular case). – Alex Salauyou Mar 10 '16 at 10:32
@Bathsheba and I think you agree that for using `==` with doubles one must deeply understand their internal structure. – Alex Salauyou Mar 10 '16 at 10:34
1

Absolutely. Your approach would be better (even upvoteble) if the 'delta' was linear on the radius of the circle. – Bathsheba Mar 10 '16 at 10:38
Given distance and radius will be positive, why the `Math.Abs`? – weston Mar 10 '16 at 11:03
@weston if `distance < radius` => `distance - radius < 0` making statement true for any `DELTA >= 0` – Alex Salauyou Mar 10 '16 at 11:12

score 0 · Answer 3 · answered Mar 10 '16 at 11:04

Are you sure this question requires doubles as input? The examples given are integers. With integers you can be sure of a points location, with real numbers (doubles) you cannot be sure about the "on circle" or not which is another reason I think the question expects you to use integers.

And the trick for performance and accuracy is to not use Math.sqrt and only work with integers.

int x;
int y;
int centreX;
int centreY;

int deltaX = x - centreX;
int deltaY = y - centreY;

int distanceSquared = deltaX * deltaX + deltaY * deltaY;

int radiusSquared = radius * radius;

if (distanceSquared == radiusSquared) { //distance == radius
  //on circle
} else if (distanceSquared < radiusSquared) { //distance < radius
  //in circle
} else {
  //out of circle
}

Point within and on circle

3 Answers3