We Keep Coding

Considering only 3D points, wouldn't this naive approach to the "smallest enclosing ball" be good enough?

Lets use C# in our example.
public class Sphere
{
public Point Center { get; set; }
public float Radius { get; set; }
public Sphere(IEnumerable<Point> points)
{
Point first = points.First();
Point vecMaxZ = first;
Point vecMinZ = first;
Point vecMaxY = first;
Point vecMinY = first;
Point vecMinX = first;
Point vecMaxX = first;
foreach (Point current in points)
{
if (current.X < vecMinX.X)
{
vecMinX = current;
}
if (current.X > vecMaxX.X)
{
vecMaxX = current;
}
if (current.Y < vecMinY.Y)
{
vecMinY = current;
}
if (current.Y > vecMaxY.Y)
{
vecMaxY = current;
}
if (current.Z < vecMinZ.Z)
{
vecMinZ = current;
}
if (current.Z > vecMaxZ.Z)
{
vecMaxZ = current;
}
}
//the lines bellow assure at least 2 points sit on the surface of the sphere.
//I'm pretty sure the algorithm is solid so far, unless I messed up the if/elses.
//I've been over this, looking at the variables and the if/elses and they all
//seem correct, but our own errors are the hardest to spot,
//so maybe there's something wrong here.
float diameterCandidateX = vecMinX.Distance(vecMaxX);
float diameterCandidateY = vecMinY.Distance(vecMaxY);
float diameterCandidateZ = vecMinZ.Distance(vecMaxZ);
Point c;
float r;
if (diameterCandidateX > diameterCandidateY)
{
if (diameterCandidateX > diameterCandidateZ)
{
c = vecMinX.Midpoint(vecMaxX);
r = diameterCandidateX / 2f;
}
else
{
c = vecMinZ.Midpoint(vecMaxZ);
r = diameterCandidateZ / 2f;
}
}
else if (diameterCandidateY > diameterCandidateZ)
{
c = vecMinY.Midpoint(vecMaxY);
r = diameterCandidateY / 2f;
}
else
{
c = vecMinZ.Midpoint(vecMaxZ);
r = diameterCandidateZ / 2f;
}
//the lines bellow look for points outside the sphere, and if one is found, then:
//1 let dist be the distance from the stray point to the current center
//2 let diff be the equal to dist - radius
//3 radius will then the increased by half of diff.
//4 a vector with the same direction as the stray point but with magnitude equal to diff is found
//5 the current center is moved by half the vector found in the step above.
//
//the stray point will now be included
//and, I would expect, the relationship between the center and other points will be mantained:
//if distance from p to center = r / k,
//then new distance from p to center' = r' / k,
//where k doesn't change from one equation to the other.
//this is where I'm wrong. I cannot figure out how to mantain this relationship.
//clearly, I'm moving the center by the wrong amount, and increasing the radius wrongly too.
//I've been over this problem for so much time, I cannot think outside the box.
//my whole world is the box. The box and I are one.
//maybe someone from outside my world (the box) could tell me where my math is wrong, please.
foreach (Point current in points)
{
float dist = current.Distance(c);
if (dist > r)
{
float diff = dist - r;
r += diff / 2f;
float scaleFactor = diff / current.Length();
Point adjust = current * scaleFactor;
c += adjust / 2f;
}
}
Center = c;
Radius = r;
}
public bool Contains(Point point) => Center.Distance(point) <= Radius;
public override string ToString() => $"Center: {Center}; Radius: {Radius}";
}
public class Point
{
public float X { get; set; }
public float Y { get; set; }
public float Z { get; set; }
public Point(float x, float y, float z)
{
X = x;
Y = y;
Z = z;
}
public float LengthSquared() => X * X + Y * Y + Z * Z;
public float Length() => (float) Math.Sqrt(X * X + Y * Y + Z * Z);
public float Distance(Point another)
{
return (float) Math.Sqrt(
(X - another.X) * (X - another.X)
+ (Y - another.Y) * (Y - another.Y)
+ (Z - another.Z) * (Z - another.Z));
}
public float DistanceSquared(Point another)
{
return (X - another.X) * (X - another.X)
+ (Y - another.Y) * (Y - another.Y)
+ (Z - another.Z) * (Z - another.Z);
}
public Point Perpendicular()
{
return new Point(-Y, X, Z);
}
public Point Midpoint(Point another)
{
return new Point(
(X + another.X) / 2f,
(Y + another.Y) / 2f,
(Z + another.Z) / 2f);
}
public override string ToString() => $"({X}, {Y}, {Z})";
public static Point operator +(Point p1, Point p2)
{
return new Point(p1.X + p2.X, p1.Y + p2.Y, p1.Z + p2.Z);
}
public static Point operator *(Point p1, float v)
{
return new Point(p1.X * v, p1.Y * v, p1.Z * v);
}
public static Point operator /(Point p1, float v)
{
return new Point(p1.X / v, p1.Y / v, p1.Z / v);
}
}
//Note: this class is here so I can be able to solve the problems suggested by
//Eric Lippert.
public class Line
{
private float coefficient;
private float constant;
public Line(Point p1, Point p2)
{
float deltaY = p2.Y - p1.Y;
float deltaX = p2.X - p1.X;
coefficient = deltaY / deltaX;
constant = coefficient * -p1.X + p1.Y;
}
public Point FromX(float x)
{
return new Point(x, x * coefficient + constant, 0);
}
public Point FromY(float y)
{
return new Point((y - constant) / coefficient, y, 0);
}
public Point Intersection(Line another)
{
float x = (another.constant - constant) / (coefficient - another.coefficient);
float y = FromX(x).Y;
return new Point(x, y, 0);
}
}
Can I safely assume this will run at least just as fast as the fancy algorithms out there that usually consider, for robustness sake, the possibility of the Points having any number of dimensions, from 2 to anything, like 1000 or 10,000 dimensions.
I only need it for 3 dimensions, never more and never less than that. Since I have no academic degree on computer science (or any degree for that matter, I'm a highschool sophomore), I have difficulties in analyzing algorithms for performance and resource consumption. So my question basically is: Is my "smallest enclosing sphere for dumbs" algoritm good in performance and resource consumption when compared with the fancy ones? Is there a point where my algorithm breaks while the professional ones don't, meaning it performs so bad it will cause noticeable loss (like, if I have too many points).
EDIT 1: I editted the code because it made no sense at all (I was hungry, it was 4pm and I haven't eaten all day). This one makes more sense I think, not sure if it's correct though. The original question stands: If this one solves the problem, does it do it well enough to compete with the stardard professional algorithms in case we know in advance that all points have 3 dimensions?
EDIT 2: Now I'm pretty sure the performance is bad, and I lost all hope of implementing a naive algorithm to find the smallest enclosing sphere. I just want to make something that work. Please, check the latest update.
EDIT 3: Doesn't work either. I quit.
EDIT 4: Finally, after, I don't know... some 5 hours. I figured it out. Jesus Christ. This one works. Could someone tell me about the performance issue? Is it really bad compared to the professional algorithms? What lines can I change to make it better? Is there a point where it breaks? Remember, I will always use it for 3D points.
EDIT 5: I learned from Bychenko the previous algorithm still didn't work. I slept on this issue, and this is my new version of the algorithm. I know it doesn't work, and I have a good clue where it is wrong, could anyone please tell why those particular calculations are wrong and how to fix them? I'm inclined to think this has something to do with trigonometry. My assumptions don't hold true for Euclidean space, because I can't stop seeing vectors as real numbers instead
of sets of real numbers that, in my case, I use to pin-point a location in Euclidean space. I'm pretty sure I'm missing some sine or cosine somewhere in the last loop (of course, not exactly sine or cosine, but the equivalent in cartesian coordinates, since we don't know any angles.
Addendum to EDIT 5: About the problems proposed by Eric Lippert:
(1) argh too trivial :p
(2) I will do it for the circle first; I will add a class Line for that.
Point a, b, c; //they are not collinear
Point midOfAB = a.Midpoint(b);
Point midOfBC = b.Midpoint(c);
//multiplying the vector by a scalar as I do bellow doesn't matter right?
Point perpendicularToAB = midOfAB.Perpendicular() * 3;
Point perpendicularToBC = midOfBC.Perpendicular() * 3;
Line bisectorAB = new Line(perpendicularToAB, midOfAB);
Line bisectorBC = new Line(perpendicularToBC, midOfBC);
Point center = bisectorAB.Intersection(bisectorBC);
float distA = center.Distance(a);
float distB = center.Distance(b);
float distC = center.Distance(c);
if(distA == distB && distB == distC)
//it works (spoiler alert: it doesn't)
else
//you're a failure, programmer, pick up your skate and practice some ollies

Sorry, but your algorithm is wrong. It doesn't solve the problem.
Counter example (3 points):
A = (0, 0, 0) - closest to origin (0)
B = (3, 3, 0) - farthest from origin (3 * sqrt(2) == 4.2426...)
C = (4, 0, 0)
your naive algorithm declares that the sphere has center at
P = (3 / sqrt(2), 3 / sqrt(2), 0)
and radius
R = 3 / sqrt(2)
and you can see that the point C = (4, 0, 0) is beyond the sphere
Edit the updated (but naive) algorithm is still wrong.
Counter example (3 points):
A = (0, 0, 0)
B = (1, 2, 0)
C = (4, 1, 0)
according the algorithm the sphere has its center at
P = (2, 1, 0)
with radius
R = sqrt(5)
and you can see that the sphere is not a minimal (smallest) one.
Nth Edit you still have an incorrect algorithm. When exploring gray zone (you know the problem, but partially, with holes) it's a good practice to invest into testing automatition. As you should know, in case of triangle all the vertexes should be on the sphere; let's validate your the solution on this fact:
public static class SphereValidator {
private static Random m_Random = new Random();
private static String Validate() {
var triangle = Enumerable
.Range(0, 3)
.Select(i => new Point(m_Random.Next(100), m_Random.Next(100), m_Random.Next(100)))
.ToArray();
Sphere solution = new Sphere(triangle);
double tolerance = 1.0e-5;
for (int i = 0; i < triangle.Length; ++i) {
double r = triangle[i].Distance(solution.Center);
if (Math.Abs(r - solution.Radius) > tolerance) {
return String.Format("Counter example\r\n A: {0}\r\n B: {1}\r\n C: {2}\r\n expected distance to \"{3}\": {4}; actual R {5}",
triangle[0], triangle[1], triangle[2], (char) ('A' + i), r, solution.Radius);
}
}
return null;
}
public static String FindCounterExample(int attempts = 10000) {
for (int i = 0; i < attempts; ++i) {
String result = Validate();
if (!String.IsNullOrEmpty(result))
Console.WriteLine(result);
return;
}
Console.WriteLine(String.Format("Yes! All {0} tests passed!", attempts));
}
}
I've just run the code above and got:
Counter example
A: (3, 30, 9)
B: (1, 63, 40)
C: (69, 1, 16)
expected distance to "A": 35.120849609375; actual R 53.62698

For a crude approximation, compute the Axis-Aligned Bounding Box, then the bounding sphere of that box (same center, diameter = √(W² + H² + D²) ).
You can refine by computing the largest distance from that center.

Calculating temporary coordinate if distance exceeds 100

I have a problem with coding a function that will do the following:
Function will take 2 sets of coordinates (x,y,z). First set is starting point and second one is end point.
Now first i will have to determine distance between those two points. If distance will be greater than lets say 100 then i will have to calculate a temporary point. I will then calculate distance between first set of coordinates and this temporary point. Please check illustration below i am sure things will be much more understandable.
Function that calculates new point between points A,B:
x = (x1+x2/2)
y = (y1+y2/2)
z = (z1+z2/2)
Function that calculates distance is:
public float DistanceTo(float x, float y, float z, float x2, float y2, float z2)
{
float a = x - x2;
float b = y - y2;
float c = z - z2;
return Math.Sqrt(a * a + b * b + c * c);
}
Link to illustration:
Perhaps my approach will be CPU heavy and perhaps slow (?) but currently i am out of ideas how to approach the problem.
Basically i need a function that will go from A to B in increments if (distance < 100).
Thank you for reading and thanks for any solution posted!
PS please dont make fun of my paint skills :)

Compute the distance d from (xb,yb,zb) (begin) to (xe,ye,ze) (end). Compute the number of parts N=(int)((d-1)/100)+1. Compute
xk = xb + (k*(xe-xb))/N
yk = yb + (k*(ye-yb))/N
zk = zb + (k*(ze-zb))/N
for k=1,...,N-1 to get equally spaced points on the segment with a distance not surpassing 100.

You can do it recursively
create a Point struct that represent a point in R3, with 2 utility methods to calculate the distance and midpoint.
struct Point
{
public double x;
public double y;
public double z;
public double Distance(Point b)
{
return Math.Sqrt(Math.Pow(b.x - this.x, 2) +
Math.Pow(b.y - this.y, 2) +
Math.Pow(b.z - this.z, 2));
}
public Point MidPoint(Point b)
{
return new Point()
{
x = (this.x + b.x) / 2,
y = (this.y + b.y) / 2,
z = (this.z + b.z) / 2
};
}
}
and write a simple recursive function that will calculate the distance and call itself recursively until the distance between point a and the temp point is less than maxSegmentLength which is 100 in your case:
public static Point GetPoint(Point a, Point b, double maxSegmentLength)
{
var distance = a.Distance(b);
if (distance < maxSegmentLength)
return b;
else
return GetPoint(a, a.MidPoint(b),maxSegmentLength);
}
Or more efficiently with vectors:
struct Vector
{
public double Vx;
public double Vy;
public double Vz;
public double R;
public Vector(Point a,Point b)
{
R = a.Distance(b);
Vx = (b.x - a.x)/R;
Vy = (b.y - a.y)/R;
Vz = (b.z - a.z)/R;
}
}
public static Point GetPoint(Point a, Point b,double maxSegmentLength)
{
var Vab = new Vector(a, b);
var dAC = Vab.R;
while (dAC > maxSegmentLength) { dAC /= 2; } //or replace this line and the one above it with var dAC=Math.Pow(0.5,(int)(-(Math.Log(maxSegmentLength / Vab.R) / Math.Log(2))) + 1)*Vab.R;
return new Point() {
x = a.x + Vab.Vx * dAC ,
y = a.y + Vab.Vy * dAC ,
z = a.z + Vab.Vz * dAC
};
}

If the point is always at a midpoint then its distance is always half of the original
public double Calc(Point A, Point B)
{
double d = Distance(A,B);
while(d>100)
{
d/=2;
}
return d;
}

Clockwise right hand turn angle in 2D?

If I have two points p1 and p2 where p1 is the pivot point and p2 is the original direction the user was headed and they have a number of possible directions to go p3...pn in random sequence. How do I get the angles between the choices and the segment formed by p1,p2 as clockwise(right hand) positive values between 0 and 360 so that I can sort them from least to greatest?
Also the points p1...pn will be in any quadrant, I can’t assume they will always be in the positive x,y direction. The grid is a standard Cartesian grid not screen coordinates so Y gets smaller as you go down not larger.
So in this example (sorry for the poor drawing but Paint was all I had on my laptop) I need to get the angles:
(p2-p1-p3)
( p2-p1-p4)
( p2-p1-p5)
( p2-p1-p6)
In this order(smallest right hand turn to largest right hand turn):
[( p2-p1-p4), ( p2-p1-p6), ( p2-p1-p5), (p2-p1-p3)]
The points in my case are a class called Vertex:
public class Vertex
{
public double X = 0;
public double Y = 0;
public Vertex() { }
public Vertex(double x, double y)
{
X = x;
Y = y;
}
}
And the code for getting the angles and sorting looks like this right now but has a problem:
private static IEnumerable<Vertex> SortByAngle(Vertex original, Vertex pivot, List<Vertex> choices)
{
choices.Sort((v1, v2) => GetTurnAngle(original, pivot, v1).CompareTo(GetTurnAngle(original, pivot, v2)));
return choices;
}
private static double GetTurnAngle(Vertex original, Vertex pivot, Vertex choice)
{
var a = original.X - pivot.X;
var b = original.Y - pivot.Y;
var c = choice.X - pivot.X;
var d = choice.Y - pivot.Y;
var rads = Math.Acos(((a * c) + (b * d)) / ((Math.Sqrt(a * a + b * b)) * (Math.Sqrt(c * c + d * d))));
return (180 / Math.PI * rads);
}
The problem is the above is if I check it for:
original 66,-66
pivot 280,-191
choice 200,-180
I get an angle of 22.460643124 instead of 337.539356876 which means it went counter-clockwise from the original direction to get that angle. I need it to always go clockwise to get the angle.
What am I doing wrong and how do I fix it?
Update: OK so according to what you guys are saying I can probably use some cross product like math to determine CW vs CCW so the new method would look like this:
private static double GetTurnAngle(Vertex original, Vertex pivot, Vertex choice)
{
var a = original.X - pivot.X;
var b = original.Y - pivot.Y;
var c = choice.X - pivot.X;
var d = choice.Y - pivot.Y;
var angle = Math.Acos(((a * c) + (b * d)) / ((Math.Sqrt(a * a + b * b)) * (Math.Sqrt(c * c + d * d))));
angle = (180 / Math.PI * angle);
var z = (choice.X - pivot.X) * (original.Y - pivot.Y) - (choice.Y - pivot.Y) * (original.X - pivot.X);
if (z < 0)
{
return 360 - angle;
}
return angle;
}
Update 2:
Using the accepted solution it now looks like so:
private static double GetTurnAngle(Vertex original, Vertex pivot, Vertex choice)
{
var angle1 = Math.Atan2(original.Y - pivot.Y, original.X - pivot.X);
var angle2 = Math.Atan2(choice.Y - pivot.Y, choice.X - pivot.X);
var angleDiff = (180 / Math.PI * (angle2 - angle1));
if (angleDiff > 0)//It went CCW so adjust
{
return 360 - angleDiff;
}
return -angleDiff;//I need the results to be always positive so flip sign
}
So far as I can tell that works great so far. Thank you guys for the help!

Take a look at atan2 function. It takes delta y and delta x, so can distinguish all angles.
angle1 = atan2(p1.y-p0.y, p1.x-p0.x);
angle2 = atan2(p2.y-p0.y, p2.x-p0.x);
angle = angle2 - angle1;
If angle is negative, then CW, if positive CCW (or other way around depending on your axis orientation). Note |angle| may be > 180, in which case you may want to do 360-|angle| and reverse the CW CCW conclusion if you're after the shortest route.

You find the Dn=direction from p1 to pn (x=pn.x-p1.x and y=pn.y-p1.y) by the formula:
Dn=f(x,y)=180-90*(1+sign(y))* (1-sign(x^2))-45*(2+sign(y))*sign(x)
-180/pi()*sign(x*y)*atan((abs(y)-abs(x))/(abs(y)+abs(x)))
So the angles are Angle(p2-p1-pn)=Dn-D2.

finding height on a heightmap stretched over a sphere C#

I'm looking for a bit of math help. I have a game were a 2D heightmap is generated and then stretched over a sphere using a length/direction formula. Now I need to know how to calculate the height between 2 points on my heightmap.
What I know:
The array that holds the heightmap
The angle in radians to my object
how many points there are on the heightmap
My problem look somewhat like so:
image
more images
The red and blue lines are the 2 heightmap points, and the light blue is where I'd like to calculate the height at.
Here's my current code to do it, but it doesn't work to well.
public double getheight(double angle)
{
//find out angle between 2 heightmap point
double offset = MathHelper.TwoPi / (heightmap.Length - 1);
//total brainfart attempt
double lowerAngle = offset * angle;
double upperAngle = offset * angle + offset;
//find heights
double height1 = heightmap[(int)lowerAngle];
double height2 = heightmap[(int)upperAngle];
//find offset angle
double u = angle - lowerAngle / (upperAngle - lowerAngle);
//return the height
return height1 + (height1 - height2) * u;
}
from my vegetation code, this seems to work okay, but is to rough to use for units and such, as they jump up/down as they move, due to it using only 1 heightmap point.
double[] hMap = planet.getHeightMap();
double i = hMap.Length / (Math.PI * 2);
this.height = hMap[(int)(angle * i)];

EDIT: example at end based on additional question info
Sounds to me like a linear interpolation - if you look at it from a 2d point of view, you've got two points:
(x1, y1) = point one on heightmap
(x2, y2) = point two on heightmap
and one point somewhere between (x1,x2) at an unknown height:
pu = (xu, yu)
A generic formula for LERP is:
pu = p0 + (p1 - p0) * u
where:
p0 = first value
p1 = second value
u = % where your unknown point lies between (p0,p1)
Here, we'll say p0 == y2 and p1 == y1. Now we need to determine "how far" the unknown point is between x1 and x2 - if you know the angles to the two heightmap points, this is easy:
u = ang(xu) - ang(x1) / (ang(x2) - ang(x1))
Alternatively, you could project your angle out to Max(y1,y2) and get the "unknown x pos" that way, then calculate the above.
So, let's try a contrived example:
p1 = point one in map = (1,2) therefore ang(p1) ~ 57 degrees
p2 = point two in map = (2,4) therefore ang(p2) ~ 114 degrees
note that here, the "x axis" is along the surface of the sphere, and the "y-axis" is the distance away from the center.
pu = object location = py #angle 100 degrees ~ 1.74 radians
px = (1.74 rad - 1 rad ) / (2 rad - 1 rad) = 0.74 / 1.0 = 0.74 => 74%
py = y0 + (y1 - y0) * u
= 2 + (4 - 2) * 0.74
= 2.96
Hopefully I didn't drop or misplace a sign there somewhere... :)
Ok, your example code - I've tweaked it a bit, here's what I've come up with:
First, let's define some helpers of my own:
public static class MathHelper
{
public const double TwoPi = Math.PI * 2.0;
public static double DegToRad(double deg)
{
return (TwoPi / 360.0) * deg;
}
public static double RadToDeg(double rad)
{
return (360.0 / TwoPi) * rad;
}
// given an upper/lower bounds, "clamp" the value into that
// range, wrapping over to lower if higher than upper, and
// vice versa
public static int WrapClamp(int value, int lower, int upper)
{
return value > upper ? value - upper - 1
: value < lower ? upper - value - 1
: value;
}
}
Our Test setup:
void Main()
{
var random = new Random();
// "sea level"
var baseDiameter = 10;
// very chaotic heightmap
heightmap = Enumerable
.Range(0, 360)
.Select(_ => random.NextDouble() * baseDiameter)
.ToArray();
// let's walk by half degrees, since that's roughly how many points we have
for(double i=0;i<360;i+=0.5)
{
var angleInDegrees = i;
var angleInRads = MathHelper.DegToRad(i);
Console.WriteLine("Height at angle {0}°({1} rad):{2} (using getheight:{3})",
angleInDegrees,
angleInRads,
heightmap[(int)angleInDegrees],
getheight(angleInRads));
}
}
double[] heightmap;
And our "getheight" method:
// assume: input angle is in radians
public double getheight(double angle)
{
//find out angle between 2 heightmap point
double dTheta = MathHelper.TwoPi / (heightmap.Length);
// our "offset" will be how many dThetas we are
double offset = angle / dTheta;
// Figure out two reference points in heightmap
// THESE MAY BE THE SAME POINT, if angle ends up
// landing on a heightmap index!
int lowerAngle = (int)offset;
int upperAngle = (int)Math.Round(
offset,
0,
MidpointRounding.AwayFromZero);
// find closest heightmap points to angle, wrapping
// around if we go under 0 or over max
int closestPointIndex = MathHelper.WrapClamp(
lowerAngle,
0,
heightmap.Length-1);
int nextPointIndex = MathHelper.WrapClamp(
upperAngle,
0,
heightmap.Length-1);
//find heights
double height1 = heightmap[closestPointIndex];
double height2 = heightmap[nextPointIndex];
// percent is (distance from angle to closest angle) / (angle "step" per heightmap point)
double percent = (angle - (closestPointIndex * dTheta)) / dTheta;
// find lerp height = firstvalue + (diff between values) * percent
double lerp = Math.Abs(height1 + (height2 - height1) * percent);
// Show what we're doing
Console.WriteLine("Delta ang:{0:f3}, Offset={1:f3} => compare indices:[{2}, {3}]",
dTheta,
offset,
closestPointIndex,
nextPointIndex);
Console.WriteLine("Lerping {0:p} between heights {1:f4} and {2:f4} - lerped height:{3:f4}",
percent,
height1,
height2,
lerp);
return lerp;
}

Code blocks because it is also trying to generate a circle around it when is a space near

I am developing a small program, and to generate the coordenates I am using this code:
public double[] GenerateNewCircle(int maxSize, int numberOfBalls, double centerx, double centery)
{
double[] position = null;
double angle = 360 / 6;
double angleRad = DegreeToRadian(incrementAngle);
position = CalculatePosition(radiusX, radiusY, centerx, centery, angleRad);
return position;
}
public double[] CalculatePosition(double radiusX, double radiusY, double centerX, double centerY, double angle)
{
double[] position = new double[2];
position[0] = Math.Cos(angle) * (radiusX + centerX)*1.5;
position[1] = Math.Sin(angle) * (radiusY + centerY)*1.5;
return position;
}
To check if not overlapping or if is in the area I used this
bool circleIsAllowed(List<Circle> circles, Circle newCircle)
{
if (newCircle.x < 10 || newCircle.x > pictureBox1.Width)
return false;
if (newCircle.y < 9 || newCircle.y > pictureBox1.Height)
return false;
foreach (Circle it in circles)
{
double aX = Math.Pow(it.x - newCircle.x, 2);
double aY = Math.Pow(it.y - newCircle.y, 2);
double Dif = Math.Abs(aX - aY);
double ra1 = it.r / 2;
double ra2 = it.r / 2;
double raDif = Math.Pow(ra1 + ra2, 2);
if ((raDif + 1) > Dif) return false;
}
return true; // no existing circle overlaps
}
I have two problems right now
It blocks because is also trying to generate a circle around it, when is a space near.
How can I increase the distance between the circles? They are too close.
This circles is to draw around a circle (around it)
This is the image:

There is mistake in second code block. Try this:
double aX = Math.Pow(it.x - newCircle.x, 2);
double aY = Math.Pow(it.y - newCircle.y, 2);
double distance = Math.Pow(aX + aY, 0.5); // distance between circles centers
double ra1 = it.r / 2;
double ra2 = it.r / 2;
// is it.r diameter? if so, then ok, else div by 2 is useless
double circlesDistance = 10; // distance between circles (not centers)
if (distance < ra1 + ra2 + circlesDistance)
{
return false;
}

To compute the distance between two circle centers, you should apply the Pythagorean Theorem:
a2 + b2 = c2.
This works perfectly if we have circles (with ellipses you'd have to compute their radius along the connection between their centres which would involve some sinus and cosinus computations).
double aX = Math.Pow(it.x - newCircle.x, 2);
double aY = Math.Pow(it.y - newCircle.y, 2);
if ((aX + aY) <= Math.Pow(it.r + newCircle.r, 2)) {
return false;
}