javascript - For the point inside circle, find in which quarter it is? -

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i researched google couldn't find keywords search. ask here if algorithm , code efficient?

http://sketchtoy.com/66429941 (algorithm)

the algoritm is: have 4 points are: north, east, south , west of circle. check 4 distances (distancetonorth, distancetoeast, distancetosouth, distancetowest). , find minimum of them quarter.

here code not seem efficient me. (firstquarter north, secondquarter east , on..

note: assume mousemove inside circle.

var firstquarterx = centerx; var firstquartery = centery - radius; var secondquarterx = centerx + radius; var secondquartery = centery; var thirdquarterx = centerx; var thirdquartery = centery + radius; var fourthquarterx = centerx - radius; var fourthquartery = centery;  var distancetofirst = math.sqrt(math.pow(x-firstquarterx, 2) + math.pow(y-firstquartery, 2)); var distancetosecond = math.sqrt(math.pow(x-secondquarterx, 2) + math.pow(y-secondquartery, 2)); var distancetothird = math.sqrt(math.pow(x-thirdquarterx, 2) + math.pow(y-thirdquartery, 2)); var distancetofourth = math.sqrt(math.pow(x-fourthquarterx, 2) + math.pow(y-fourthquartery, 2));  var min = math.min(distancetofirst, distancetosecond, distancetothird, distancetofourth);  var numbers = [distancetofirst, distancetosecond, distancetothird, distancetofourth];  var index = numbers.indexof(min); // give 0 or 1 or 2 or 3  var quarter = index + 1;

observe boundaries between quarters lie along lines equations y = x , y = -x, relative origin @ center of circle. can use evaluate quarter each point falls in.

if point (x, y), coordinates relative center of circle xrelative = x - centerx , yrelative = y - centery.

  • your point in first (south in code) quarter if yrelative < 0 , math.abs(xrelative) < -yrelative
  • your point in second (east) quarter if xrelative > 0 , math.abs(yrelative) < xrelative
  • your point in third (north) quarter if yrelative > 0 , math.abs(xrelative) < yrelative
  • your point in fourth (west) quarter if xrelative < 0 , math.abs(yrelative) < -xrelative

i leave determine quarter assign points fall on boundary. also, can implement little decision tree based on criteria if prefer; should little more efficient testing each criterion in turn.


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