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I've looked around but can't find anything useful for drawing circles in golang.
I'd like to draw a draw with 2 given (inner and outer) radius and color all pixels in between.
One possible method would be to iterate through each pixel and color it until the ring has been created. Although, that seems really inefficient.
Any help on this would be greatly appreciated! :)
Please read this related question: Draw a rectangle in Golang?
To sum it up: the standard Go library does not provide primitive drawing or painting capabilities.
So yes, either you have to use a 3rd party library to draw a circle (such as github.com/llgcode/draw2d), or you have to do it yourself. Don't worry, it's not hard at all.
Drawing a single circle
First pick a circle drawing algorithm which is simple and efficient. I recommend the Midpoint circle algorithm.
You will find the algorithm on the linked Wikipedia page. Note: you do not have to understand it if you want to use it.
But we do need to implement the algorithm in Go. Which is rather simple:
func drawCircle(img draw.Image, x0, y0, r int, c color.Color) {
x, y, dx, dy := r-1, 0, 1, 1
err := dx - (r * 2)
for x > y {
img.Set(x0+x, y0+y, c)
img.Set(x0+y, y0+x, c)
img.Set(x0-y, y0+x, c)
img.Set(x0-x, y0+y, c)
img.Set(x0-x, y0-y, c)
img.Set(x0-y, y0-x, c)
img.Set(x0+y, y0-x, c)
img.Set(x0+x, y0-y, c)
if err <= 0 {
y++
err += dy
dy += 2
}
if err > 0 {
x--
dx += 2
err += dx - (r * 2)
}
}
}
That's all it takes. Just pass a draw.Image you want to draw on, and the parameters of the circle you want to draw (center point, radius and the color).
Let's see it in action. Let's create an image, draw a circle on it, and save the image to a file. This is all it takes:
img := image.NewRGBA(image.Rect(0, 0, 100, 100))
drawCircle(img, 40, 40, 30, color.RGBA{255, 0, 0, 255})
buf := &bytes.Buffer{}
if err := png.Encode(buf, img); err != nil {
panic(err)
}
if err := ioutil.WriteFile("circle.png", buf.Bytes(), 0666); err != nil {
panic(err)
}
Note: you could also encode the image directly to an os.File and "skip" the in-memory buffer. This is just for demonstration, and to verify our implementation works.
Drawing a ring (filling space between 2 circles)
This one isn't that trivial if you want to implement it yourself, but using a 3rd party lib here could really come handy.
Although most of them don't contain ring-painting support, they do have circle drawing support, and you can set the width of the line used to draw the circle.
So, set the line width to the value that is the difference of the 2 radius of your circles. And draw a circle with a new radius that is the arithmetic center of the 2 original radius.
Here's the algorithm (this is not runnable code):
// Helper functions abstracting the library you choose:
func setColor(c color.Color) {}
func setLineWidth(width float64) {}
func drawCircle(r, x, y float64) {}
// fillRing draws a ring, where r1 and r2 are 2 concentric circles,
// the boundaries of the ring, (x, y) being the center point.
func fillRing(r1, r2, x, y float64, c color.color) {
// Set drawing color:
setColor(c)
// Set line width:
width := r2 - r1
if width < 0 {
width = -width
}
setLineWidth(width)
// And finally draw a circle which will be a ring:
r := (r2 + r1) / 2
drawCircle(r, x, y)
}
I have a terrain and an object which should moves over the terrain so I made a function that can detect which face is the origin of the object locate above so I should set the Y of the object with the accurate height of the intersection point between the vector from the object origin perpendicular with the face, I have the three vertices of the face so I can calculate its normal and its origin and maximum and minimum pointes (bounded box).
enter image description here
Edit:
the question in another form:
if I know x and z of a point over a triangle in the 3D space how can I know it y if it is a point in the triangle?
the Möller–Trumbore intersection algorithm
this is the implementation:
bool EngineItem::checkIntersection(glm::vec3& rayOrigin, glm::vec3& rayVector, Face& face, glm::vec3& point) {
const float EPSILON = 0.0000001;
glm::vec3 vertex0 = face.v1;
glm::vec3 vertex1 = face.v2;
glm::vec3 vertex2 = face.v3;
glm::vec3 edge1, edge2, h, s, q;
float a, f, u, v;
edge1 = vertex1 - vertex0;
edge2 = vertex2 - vertex0;
h = glm::cross(rayVector, edge2);
a = glm::dot(edge1, h);
if (a > -EPSILON && a < EPSILON)
return false;
f = 1 / a;
s = rayOrigin - vertex0;
u = f * glm::dot(s, h);
if (u < 0.0 || u > 1.0)
return false;
q = glm::cross(s, edge1);
v = f * glm::dot(rayVector, q);
if (v < 0.0 || u + v > 1.0)
return false;
float t = f * glm::dot(edge2, q);
if (t > EPSILON) {
glm::vec3 plus = glm::normalize(rayVector) * (t * glm::length(rayVector));
point = rayOrigin + plus;
return true;
} else
return false;
}
I need an algorithm to give me coordinates to the nearest cells (in order of distance) to another cell in a 2D grid. Its for a search algorithm that then checks those coordinates for all sorts of things for suitability. Anyways, so far I came up with this:
function testy(cx, cy, idx) {
var radius = Math.floor(Math.sqrt(idx / Math.PI));
var segment = Math.round(idx - (radius * Math.PI));
var angle = segment / radius;
var x = Math.round(cx + radius * Math.cos(angle));
var y = Math.round(cy + radius * Math.sin(angle));
return [x, y];
}
addEventListener("load", function() {
var canv = document.createElement("canvas");
document.body.appendChild(canv);
canv.width = 800;
canv.height = 600;
var ctx = canv.getContext("2d");
var scale = 5;
var idx = 0;
var idx_end = 10000;
var func = function() {
var xy = testy(0,0,idx++);
var x = xy[0] * scale + canv.width / 2;
var y = xy[1] * scale + canv.height / 2;
ctx.rect(x, y, scale, scale);
ctx.fill();
if (idx < idx_end) setTimeout(func, 0);
}
func();
});
but as you can tell, its kinda crap because it skips some cells. There's a few assumptions I'm making there:
That the circumference of a circle of a certain radius corresponds to the number of cells on the path of that circle. I didn't think that would be too great of a problem though since the actual number of cells in a radius should be lower than the circumference leading to duplication(which in small amounts is ok) but not exclusion(not ok).
That the radius of a circle by the n-th index specified would be slightly more than Math.floor(Math.sqrt(idx / Math.PI)) because each increase of 1 to the radius corresponds to 2 * Math.PI being added to the circumference of the circle. Again, should lead to slight duplication but no exclusion.
Other than that I have no idea what could be wrong with it, I fail at math any more complex than this so probably something to do with that.
Perhaps there is another algorithm like this already out there though? One that doesn't skip cells? Language doesn't really matter, I'm using js to prototype it but it can be whatever.
Instead of thinking about the full circle, think about a quadrant. Adapting that to the full circle later should be fairly easy. Use (0,0) as the center of the circle for convenience. So you want to list grid cells with x,y ≥ 0 in order of non-decreasing x² + y².
One useful data structure is a priority queue. It can be used to keep track of the next y value for every x value, and you can extract the one with minimal x² + y² easily.
q = empty priority queue, for easy access to element with minimal x²+y²
Insert (0,0) into queue
while queue is not empty:
remove minimal element from queue and call it (x,y)
insert (x,y+1) into queue unless y+1 is off canvas
if y = 0:
insert (x+1,0) into queue unless x+1 is off canvas
do whatever you want to do with (x,y)
So for a canvas of size n this will enumerate all the n² points, but the priority queue will only contain n elements at most. The whole loop runs in O(n² log(n)). And if you abort the loop eraly because you found what you were looking for, it gets cheaper still, in contrast to simply sorting all the points. Another benefit is that you can use integer arithmetic exclusively, so numeric errors won't be an issue. One drawback is that JavaScript does not come with a priority queue out of the box, but I'm sure you can find an implementation you can reuse, e.g. tiniqueue.
When doing full circle, you'd generate (−x,y) unless x=0, and likewise for (x,−y) and (−x,−y). You could exploit symmetry a bit more by only having the loop over ⅛ of the circle, i.e. not inserting (x,y+1) if x=y, and then also generating (y,x) as a separate point unless x=y. Difference in performance should be marginal for many use cases.
"use strict";
function distCompare(a, b) {
const a2 = a.x*a.x + a.y*a.y;
const b2 = b.x*b.x + b.y*b.y;
return a2 < b2 ? -1 : a2 > b2 ? 1 : 0;
}
// Yields points in the range -w <= x <= w and -h <= y <= h
function* aroundOrigin(w,h) {
const q = TinyQueue([{x:0, y:0}], distCompare);
while (q.length) {
const p = q.pop();
yield p;
if (p.x) yield {x:-p.x, y:p.y};
if (p.y) yield {x:p.x, y:-p.y};
if (p.x && p.y) yield {x:-p.x, y:-p.y};
if (p.y < h) q.push({x:p.x, y:p.y+1});
if (p.y == 0 && p.x < w) q.push({x:p.x + 1, y:0});
}
}
// Yields points around (cx,cy) in range 0 <= x < w and 0 <= y < h
function* withOffset(cx, cy, w, h) {
const delegate = aroundOrigin(
Math.max(cx, w - cx - 1), Math.max(cy, h - cy - 1));
for(let p of delegate) {
p = {x: p.x + cx, y: p.y + cy};
if (p.x >= 0 && p.x < w && p.y >= 0 && p.y < h) yield p;
}
}
addEventListener("load", function() {
const canv = document.createElement("canvas");
document.body.appendChild(canv);
const cw = 800, ch = 600;
canv.width = cw;
canv.height = ch;
const ctx = canv.getContext("2d");
const scale = 5;
const w = Math.ceil(cw / scale);
const h = Math.ceil(ch / scale);
const cx = w >> 1, cy = h >> 1;
const pointgen = withOffset(cx, cy, w, h);
let cntr = 0;
var func = function() {
const {value, done} = pointgen.next();
if (done) return;
if (cntr++ % 16 === 0) {
// lighten older parts so that recent activity is more visible
ctx.fillStyle = "rgba(255,255,255,0.01)";
ctx.fillRect(0, 0, cw, ch);
ctx.fillStyle = "rgb(0,0,0)";
}
ctx.fillRect(value.x * scale, value.y*scale, scale, scale);
setTimeout(func, 0);
}
func();
});
<script type="text/javascript">module={};</script>
<script src="https://cdn.rawgit.com/mourner/tinyqueue/54dc3eb1/index.js"></script>
I have a set of 2D Cartesian points [b], which proceed clockwise from the start and form a closed shape. Each one of these has its own companion 2D Cartesian points q0 and q1 which define the Beziér curve around the point (along with the preceding and succeeding points). Together, all these points define a closed 2D composite Beziér curve.
I have a separate point p which is an arbitrary 2D Cartesian point on the same plane. Is there a simple algorithm for finding the (x, y) coordinates of a new 2D Cartesian point q which is the closest point on the path to p?
As illustrated here, I have the points b[0] to b[3] and their handles b[n].q0 and b[n].q1, and I have the arbitrary point p. I'm trying to calculate the point q, not as a floating-point position along the curve, but as a pair of (x, y) coordinates.
I tried searching this, but some seemed to only be for a very small curve, others were way over my head with abstract mathematics and scientific research papers.
Any help that leads me toward an algorithmic solution is greatly appreciated, especially if it can be translated into a C-like language rather than the pure math in the above SO answers.
By adapting the algorithm posted by Tatarize, I came up with this solution in Swift, which should be translatable to other languages:
struct BezierPoint {
let q0: CGPoint
let point: CGPoint
let q1: CGPoint
}
struct SimpleBezierCurve {
let left: BezierPoint
let right: BezierPoint
}
class BezierPath {
var pathPoints = [BezierPoint]()
func findClosestPoint(to targetPoint: CGPoint) -> CGPoint {
let segments = allSegments()
guard segments.count > 0 else { return targetPoint }
var closestPoint = (distance: CGFloat.infinity, point: CGPoint(x: CGFloat.infinity, y: CGFloat.infinity))
segments.forEach{ curve in
let thisPoint = BezierPath.findClosestPoint(to: targetPoint, along: curve)
let distance = findDistance(from: targetPoint, to: thisPoint)
if (distance < closestPoint.distance) {
closestPoint = (distance: distance, point: thisPoint)
}
}
return closestPoint.point
}
func allSegments() -> [SimpleBezierCurve] {
guard pathPoints.count > 0 else { return [] }
var segments = [SimpleBezierCurve]()
var prevPoint = pathPoints[0]
for i in 1 ..< pathPoints.count {
let thisPoint = pathPoints[i]
segments.append(SimpleBezierCurve(left: prevPoint, right: thisPoint))
prevPoint = thisPoint
}
segments.append(SimpleBezierCurve(left: prevPoint, right: pathPoints[0]))
return segments
}
static func findClosestPoint(to point: CGPoint, along curve: SimpleBezierCurve) -> CGPoint {
return findClosestPointToCubicBezier(to: point, slices: 10, iterations: 10, along: curve)
}
// Adapted from https://stackoverflow.com/a/34520607/3939277
static func findClosestPointToCubicBezier(to target: CGPoint, slices: Int, iterations: Int, along curve: SimpleBezierCurve) -> CGPoint {
return findClosestPointToCubicBezier(iterations: iterations, to: target, start: 0, end: 1, slices: slices, along: curve)
}
// Adapted from https://stackoverflow.com/a/34520607/3939277
private static func findClosestPointToCubicBezier(iterations iterations: Int, to: CGPoint, start: CGFloat, end: CGFloat, slices: Int, along curve: SimpleBezierCurve) -> CGPoint {
if iterations <= 0 {
let position = (start + end) / 2
let point = self.point(for: position, along: curve)
return point
}
let tick = (end - start) / slices
var best = CGFloat(0)
var bestDistance = CGFloat.infinity
var currentDistance: CGFloat
var t = start
while (t <= end) {
//B(t) = (1-t)**3 p0 + 3(1 - t)**2 t P1 + 3(1-t)t**2 P2 + t**3 P3
let point = self.point(for: t, along: curve)
var dx = point.x - to.x;
var dy = point.y - to.y;
dx *= dx;
dy *= dy;
currentDistance = dx + dy;
if (currentDistance < bestDistance) {
bestDistance = currentDistance;
best = t;
}
t += tick;
}
return findClosestPointToCubicBezier(iterations: iterations - 1, to: to, start: max(best - tick, 0.0), end: min(best + tick, 1.0), slices: slices, along: curve);
}
static func point(for t: CGFloat, along curve: SimpleBezierCurve) -> CGPoint {
// This had to be broken up to avoid the "Expression too complex" error
let x0 = curve.left.point.x
let x1 = curve.left.q1.x
let x2 = curve.right.q0.x
let x3 = curve.right.point.x
let y0 = curve.left.point.y
let y1 = curve.left.q1.y
let y2 = curve.right.q0.y
let y3 = curve.right.point.y
let x = (1 - t) * (1 - t) * (1 - t) * x0 + 3 * (1 - t) * (1 - t) * t * x1 + 3 * (1 - t) * t * t * x2 + t * t * t * x3
let y = (1 - t) * (1 - t) * (1 - t) * y0 + 3 * (1 - t) * (1 - t) * t * y1 + 3 * (1 - t) * t * t * y2 + t * t * t * y3
return CGPoint(x: x, y: y)
}
}
// Possibly in another file
func findDistance(from a: CGPoint, to b: CGPoint) -> CGFloat {
return sqrt(pow(a.x - b.x, 2) + pow(a.y - b.y, 2));
}
GitHub Gist
If you have a circle with center (center_x, center_y) and radius radius, how do you test if a given point with coordinates (x, y) is inside the circle?
In general, x and y must satisfy (x - center_x)² + (y - center_y)² < radius².
Please note that points that satisfy the above equation with < replaced by == are considered the points on the circle, and the points that satisfy the above equation with < replaced by > are considered the outside the circle.
Mathematically, Pythagoras is probably a simple method as many have already mentioned.
(x-center_x)^2 + (y - center_y)^2 < radius^2
Computationally, there are quicker ways. Define:
dx = abs(x-center_x)
dy = abs(y-center_y)
R = radius
If a point is more likely to be outside this circle then imagine a square drawn around it such that it's sides are tangents to this circle:
if dx>R then
return false.
if dy>R then
return false.
Now imagine a square diamond drawn inside this circle such that it's vertices touch this circle:
if dx + dy <= R then
return true.
Now we have covered most of our space and only a small area of this circle remains in between our square and diamond to be tested. Here we revert to Pythagoras as above.
if dx^2 + dy^2 <= R^2 then
return true
else
return false.
If a point is more likely to be inside this circle then reverse order of first 3 steps:
if dx + dy <= R then
return true.
if dx > R then
return false.
if dy > R
then return false.
if dx^2 + dy^2 <= R^2 then
return true
else
return false.
Alternate methods imagine a square inside this circle instead of a diamond but this requires slightly more tests and calculations with no computational advantage (inner square and diamonds have identical areas):
k = R/sqrt(2)
if dx <= k and dy <= k then
return true.
Update:
For those interested in performance I implemented this method in c, and compiled with -O3.
I obtained execution times by time ./a.out
I implemented this method, a normal method and a dummy method to determine timing overhead.
Normal: 21.3s
This: 19.1s
Overhead: 16.5s
So, it seems this method is more efficient in this implementation.
// compile gcc -O3 <filename>.c
// run: time ./a.out
#include <stdio.h>
#include <stdlib.h>
#define TRUE (0==0)
#define FALSE (0==1)
#define ABS(x) (((x)<0)?(0-(x)):(x))
int xo, yo, R;
int inline inCircle( int x, int y ){ // 19.1, 19.1, 19.1
int dx = ABS(x-xo);
if ( dx > R ) return FALSE;
int dy = ABS(y-yo);
if ( dy > R ) return FALSE;
if ( dx+dy <= R ) return TRUE;
return ( dx*dx + dy*dy <= R*R );
}
int inline inCircleN( int x, int y ){ // 21.3, 21.1, 21.5
int dx = ABS(x-xo);
int dy = ABS(y-yo);
return ( dx*dx + dy*dy <= R*R );
}
int inline dummy( int x, int y ){ // 16.6, 16.5, 16.4
int dx = ABS(x-xo);
int dy = ABS(y-yo);
return FALSE;
}
#define N 1000000000
int main(){
int x, y;
xo = rand()%1000; yo = rand()%1000; R = 1;
int n = 0;
int c;
for (c=0; c<N; c++){
x = rand()%1000; y = rand()%1000;
// if ( inCircle(x,y) ){
if ( inCircleN(x,y) ){
// if ( dummy(x,y) ){
n++;
}
}
printf( "%d of %d inside circle\n", n, N);
}
You can use Pythagoras to measure the distance between your point and the centre and see if it's lower than the radius:
def in_circle(center_x, center_y, radius, x, y):
dist = math.sqrt((center_x - x) ** 2 + (center_y - y) ** 2)
return dist <= radius
EDIT (hat tip to Paul)
In practice, squaring is often much cheaper than taking the square root and since we're only interested in an ordering, we can of course forego taking the square root:
def in_circle(center_x, center_y, radius, x, y):
square_dist = (center_x - x) ** 2 + (center_y - y) ** 2
return square_dist <= radius ** 2
Also, Jason noted that <= should be replaced by < and depending on usage this may actually make sense even though I believe that it's not true in the strict mathematical sense. I stand corrected.
boolean isInRectangle(double centerX, double centerY, double radius,
double x, double y)
{
return x >= centerX - radius && x <= centerX + radius &&
y >= centerY - radius && y <= centerY + radius;
}
//test if coordinate (x, y) is within a radius from coordinate (center_x, center_y)
public boolean isPointInCircle(double centerX, double centerY,
double radius, double x, double y)
{
if(isInRectangle(centerX, centerY, radius, x, y))
{
double dx = centerX - x;
double dy = centerY - y;
dx *= dx;
dy *= dy;
double distanceSquared = dx + dy;
double radiusSquared = radius * radius;
return distanceSquared <= radiusSquared;
}
return false;
}
This is more efficient, and readable. It avoids the costly square root operation. I also added a check to determine if the point is within the bounding rectangle of the circle.
The rectangle check is unnecessary except with many points or many circles. If most points are inside circles, the bounding rectangle check will actually make things slower!
As always, be sure to consider your use case.
You should check whether the distance from the center of the circle to the point is smaller than the radius
using Python
if (x-center_x)**2 + (y-center_y)**2 <= radius**2:
# inside circle
Find the distance between the center of the circle and the points given. If the distance between them is less than the radius then the point is inside the circle.
if the distance between them is equal to the radius of the circle then the point is on the circumference of the circle.
if the distance is greater than the radius then the point is outside the circle.
int d = r^2 - ((center_x-x)^2 + (center_y-y)^2);
if(d>0)
print("inside");
else if(d==0)
print("on the circumference");
else
print("outside");
Calculate the Distance
D = Math.Sqrt(Math.Pow(center_x - x, 2) + Math.Pow(center_y - y, 2))
return D <= radius
that's in C#...convert for use in python...
As said above -- use Euclidean distance.
from math import hypot
def in_radius(c_x, c_y, r, x, y):
return math.hypot(c_x-x, c_y-y) <= r
The equation below is a expression that tests if a point is within a given circle where xP & yP are the coordinates of the point, xC & yC are the coordinates of the center of the circle and R is the radius of that given circle.
If the above expression is true then the point is within the circle.
Below is a sample implementation in C#:
public static bool IsWithinCircle(PointF pC, Point pP, Single fRadius){
return Distance(pC, pP) <= fRadius;
}
public static Single Distance(PointF p1, PointF p2){
Single dX = p1.X - p2.X;
Single dY = p1.Y - p2.Y;
Single multi = dX * dX + dY * dY;
Single dist = (Single)Math.Round((Single)Math.Sqrt(multi), 3);
return (Single)dist;
}
This is the same solution as mentioned by Jason Punyon, but it contains a pseudo-code example and some more details. I saw his answer after writing this, but I didn't want to remove mine.
I think the most easily understandable way is to first calculate the distance between the circle's center and the point. I would use this formula:
d = sqrt((circle_x - x)^2 + (circle_y - y)^2)
Then, simply compare the result of that formula, the distance (d), with the radius. If the distance (d) is less than or equal to the radius (r), the point is inside the circle (on the edge of the circle if d and r are equal).
Here is a pseudo-code example which can easily be converted to any programming language:
function is_in_circle(circle_x, circle_y, r, x, y)
{
d = sqrt((circle_x - x)^2 + (circle_y - y)^2);
return d <= r;
}
Where circle_x and circle_y is the center coordinates of the circle, r is the radius of the circle, and x and y is the coordinates of the point.
My answer in C# as a complete cut & paste (not optimized) solution:
public static bool PointIsWithinCircle(double circleRadius, double circleCenterPointX, double circleCenterPointY, double pointToCheckX, double pointToCheckY)
{
return (Math.Pow(pointToCheckX - circleCenterPointX, 2) + Math.Pow(pointToCheckY - circleCenterPointY, 2)) < (Math.Pow(circleRadius, 2));
}
Usage:
if (!PointIsWithinCircle(3, 3, 3, .5, .5)) { }
As stated previously, to show if the point is in the circle we can use the following
if ((x-center_x)^2 + (y - center_y)^2 < radius^2) {
in.circle <- "True"
} else {
in.circle <- "False"
}
To represent it graphically we can use:
plot(x, y, asp = 1, xlim = c(-1, 1), ylim = c(-1, 1), col = ifelse((x-center_x)^2 + (y - center_y)^2 < radius^2,'green','red'))
draw.circle(0, 0, 1, nv = 1000, border = NULL, col = NA, lty = 1, lwd = 1)
Moving into the world of 3D if you want to check if a 3D point is in a Unit Sphere you end up doing something similar. All that is needed to work in 2D is to use 2D vector operations.
public static bool Intersects(Vector3 point, Vector3 center, float radius)
{
Vector3 displacementToCenter = point - center;
float radiusSqr = radius * radius;
bool intersects = displacementToCenter.magnitude < radiusSqr;
return intersects;
}
iOS 15, Accepted Answer written in Swift 5.5
func isInRectangle(center: CGPoint, radius: Double, point: CGPoint) -> Bool
{
return point.x >= center.x - radius && point.x <= center.x + radius &&
point.y >= center.y - radius && point.y <= center.y + radius
}
//test if coordinate (x, y) is within a radius from coordinate (center_x, center_y)
func isPointInCircle(center: CGPoint,
radius:Double, point: CGPoint) -> Bool
{
if(isInRectangle(center: center, radius: radius, point: point))
{
var dx:Double = center.x - point.x
var dy:Double = center.y - point.y
dx *= dx
dy *= dy
let distanceSquared:Double = dx + dy
let radiusSquared:Double = radius * radius
return distanceSquared <= radiusSquared
}
return false
}
I used the code below for beginners like me :).
public class incirkel {
public static void main(String[] args) {
int x;
int y;
int middelx;
int middely;
int straal; {
// Adjust the coordinates of x and y
x = -1;
y = -2;
// Adjust the coordinates of the circle
middelx = 9;
middely = 9;
straal = 10;
{
//When x,y is within the circle the message below will be printed
if ((((middelx - x) * (middelx - x))
+ ((middely - y) * (middely - y)))
< (straal * straal)) {
System.out.println("coordinaten x,y vallen binnen cirkel");
//When x,y is NOT within the circle the error message below will be printed
} else {
System.err.println("x,y coordinaten vallen helaas buiten de cirkel");
}
}
}
}}
Here is the simple java code for solving this problem:
and the math behind it : https://math.stackexchange.com/questions/198764/how-to-know-if-a-point-is-inside-a-circle
boolean insideCircle(int[] point, int[] center, int radius) {
return (float)Math.sqrt((int)Math.pow(point[0]-center[0],2)+(int)Math.pow(point[1]-center[1],2)) <= radius;
}
PHP
if ((($x - $center_x) ** 2 + ($y - $center_y) ** 2) <= $radius **2) {
return true; // Inside
} else {
return false; // Outside
}