关于算法：圆圆交点

Circle-circle intersection points

如何计算两个圆的交点。我希望在所有情况下都可以有两个，一个或没有交点。

我有中心点的x和y坐标，以及每个圆的半径。

最好使用python的答案，但是任何有效的算法都是可以接受的。

两个圆的交点

保罗·伯克(Paul Bourke)撰写

The following note describes how to find the intersection point(s)
between two circles on a plane, the following notation is used. The
aim is to find the two points P₃ = (x₃,
y₃) if they exist.

First calculate the distance d between the center
of the circles. d = ||P₁ - P₀||.

If d > r₀ + r₁ then there are no solutions,
the circles are separate.

If d < |r₀ -
r₁| then there are no solutions because one circle is
contained within the other.

If d = 0 and r₀ =
r₁ then the circles are coincident and there are an
infinite number of solutions.

Considering the two triangles P₀P₂P₃
and P₁P₂P₃ we can write

a2 + h2 = r₀2 and
b2 + h2 = r₁2

Using d = a + b we can solve for a,

a =
(r₀2 - r₁2 +
d2 ) / (2 d)

It can be readily shown that this reduces to
r₀ when the two circles touch at one point, ie: d =
r₀ + r₁

Solve for h by substituting a into the first
equation, h2 = r₀2 - a2

So

P₂ = P₀ + a ( P₁ -
P₀ ) / d

And finally, P₃ =
(x₃,y₃) in terms of P₀ =
(x₀,y₀), P₁ =
(x₁,y₁) and P₂ =
(x₂,y₂), is

x₃ =
x₂ +- h ( y₁ - y₀ ) / d

y₃ = y₂ -+ h ( x₁ - x₀ ) /
d

资料来源：http://paulbourke.net/geometry/circlesphere/

相关讨论

这是基于Paul Bourke的文章的C ++实现。它只有在有两个交叉点时才有效，否则可能返回NaN NAN NAN NAN。

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class Point{
public:
float x, y;
Point(float px, float py) {
x = px;
y = py;
}
Point sub(Point p2) {
return Point(x - p2.x, y - p2.y);
}
Point add(Point p2) {
return Point(x + p2.x, y + p2.y);
}
float distance(Point p2) {
return sqrt((x - p2.x)*(x - p2.x) + (y - p2.y)*(y - p2.y));
}
Point normal() {
float length = sqrt(x*x + y*y);
return Point(x/length, y/length);
}
Point scale(float s) {
return Point(x*s, y*s);
}
};

class Circle {
public:
float x, y, r, left;
Circle(float cx, float cy, float cr) {
x = cx;
y = cy;
r = cr;
left = x - r;
}
pair<Point, Point> intersections(Circle c) {
Point P0(x, y);
Point P1(c.x, c.y);
float d, a, h;
d = P0.distance(P1);
a = (r*r - c.r*c.r + d*d)/(2*d);
h = sqrt(r*r - a*a);
Point P2 = P1.sub(P0).scale(a/d).add(P0);
float x3, y3, x4, y4;
x3 = P2.x + h*(P1.y - P0.y)/d;
y3 = P2.y - h*(P1.x - P0.x)/d;
x4 = P2.x - h*(P1.y - P0.y)/d;
y4 = P2.y + h*(P1.x - P0.x)/d;

return pair<Point, Point>(Point(x3, y3), Point(x4, y4));
}

};

相关讨论

这是使用向量的Javascript实现。该代码有充分的文档记录，您应该可以遵循它。这是原始来源

观看现场演示：
enter image description here

// Let EPS (epsilon) be a small value
var EPS = 0.0000001;

// Let a point be a pair: (x, y)
function Point(x, y) {
this.x = x;
this.y = y;
}

// Define a circle centered at (x,y) with radius r
function Circle(x,y,r) {
this.x = x;
this.y = y;
this.r = r;
}

// Due to double rounding precision the value passed into the Math.acos
// function may be outside its domain of [-1, +1] which would return
// the value NaN which we do not want.
function acossafe(x) {
if (x >= +1.0) return 0;
if (x <= -1.0) return Math.PI;
return Math.acos(x);
}

// Rotates a point about a fixed point at some angle 'a'
function rotatePoint(fp, pt, a) {
var x = pt.x - fp.x;
var y = pt.y - fp.y;
var xRot = x * Math.cos(a) + y * Math.sin(a);
var yRot = y * Math.cos(a) - x * Math.sin(a);
return new Point(fp.x+xRot,fp.y+yRot);
}

// Given two circles this method finds the intersection
// point(s) of the two circles (if any exists)
function circleCircleIntersectionPoints(c1, c2) {

var r, R, d, dx, dy, cx, cy, Cx, Cy;

if (c1.r < c2.r) {
r = c1.r; R = c2.r;
cx = c1.x; cy = c1.y;
Cx = c2.x; Cy = c2.y;
} else {
r = c2.r; R = c1.r;
Cx = c1.x; Cy = c1.y;
cx = c2.x; cy = c2.y;
}

// Compute the vector <dx, dy>
dx = cx - Cx;
dy = cy - Cy;

// Find the distance between two points.
d = Math.sqrt( dx*dx + dy*dy );

// There are an infinite number of solutions
// Seems appropriate to also return null
if (d < EPS && Math.abs(R-r) < EPS) return [];

// No intersection (circles centered at the
// same place with different size)
else if (d < EPS) return [];

var x = (dx / d) * R + Cx;
var y = (dy / d) * R + Cy;
var P = new Point(x, y);

// Single intersection (kissing circles)
if (Math.abs((R+r)-d) < EPS || Math.abs(R-(r+d)) < EPS) return [P];

// No intersection. Either the small circle contained within
// big circle or circles are simply disjoint.
if ( (d+r) < R || (R+r < d) ) return [];

var C = new Point(Cx, Cy);
var angle = acossafe((r*r-d*d-R*R)/(-2.0*d*R));
var pt1 = rotatePoint(C, P, +angle);
var pt2 = rotatePoint(C, P, -angle);
return [pt1, pt2];

}

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为什么不使用以下7种您喜欢的过程语言(或可编程计算器！)呢？

假设给定了P0坐标(x0，y0)，P1坐标(x1，y1)，r0和r1，并且您想找到P3坐标(x3，y3)：

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d=sqr((x1-x0)^2 + (y1-y0)^2)
a=(r0^2-r1^2+d^2)/(2*d)
h=sqr(r0^2-a^2)
x2=x0+a*(x1-x0)/d
y2=y0+a*(y1-y0)/d
x3=x2+h*(y1-y0)/d // also x3=x2-h*(y1-y0)/d
y3=y2-h*(x1-x0)/d // also y3=y2+h*(x1-x0)/d

相关讨论

尝试这个;

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def ri(cr1,cr2,cp1,cp2):
int1=[]
int2=[]
ori=0
if cp1[0]<cp2[0] and cp1[1]!=cp2[1]:
p1=cp1
p2=cp2
r1=cr1
r2=cr2
if cp1[1]<cp2[1]:
ori+=1
elif cp1[1]>cp2[1]:
ori+=2
elif cp1[0]>cp2[0] and cp1[1]!=cp2[1]:
p1=cp2
p2=cp1
r1=cr2
r2=cr1
if p1[1]<p2[1]:
ori+=1
elif p1[1]>p2[1]:
ori+=2
elif cp1[0]==cp2[0]:
ori+=4
if cp1[1]>cp2[1]:
p1=cp1
p2=cp2
r1=cr1
r2=cr2
elif cp1[1]<cp2[1]:
p1=cp2
p2=cp1
r1=cr2
r2=cr1
elif cp1[1]==cp2[1]:
ori+=3
if cp1[0]>cp2[0]:
p1=cp2
p2=cp1
r1=cr2
r2=cr1
elif cp1[0]<cp2[0]:
p1=cp1
p2=cp2
r1=cr1
r2=cr2
if ori==1:#+
D=calc_dist(p1,p2)
tr=r1+r2
el=tr-D
a=r1-el
b=r2-el
A=a+(el/2)
B=b+(el/2)
thta=math.degrees(math.acos(A/r1))
rs=p2[1]-p1[1]
rn=p2[0]-p1[0]
gd=rs/rn
yint=p1[1]-((gd)*p1[0])
dty=calc_dist(p1,[0,yint])

aa=p1[1]-yint
bb=math.degrees(math.asin(aa/dty))
d=90-bb
e=180-d-thta
g=(dty/math.sin(math.radians(e)))*math.sin(math.radians(thta))
f=(g/math.sin(math.radians(thta)))*math.sin(math.radians(d))
oty=yint+g
h=f+r1
i=90-e
j=180-90-i
l=math.sin(math.radians(i))*h
k=math.cos(math.radians(i))*h
iy2=oty-l
ix2=k
int2.append(ix2)
int2.append(iy2)

m=90+bb
n=180-m-thta
p=(dty/math.sin(math.radians(n)))*math.sin(math.radians(m))
o=(p/math.sin(math.radians(m)))*math.sin(math.radians(thta))
q=p+r1
r=90-n
s=math.sin(math.radians(r))*q
t=math.cos(math.radians(r))*q
otty=yint-o
iy1=otty+s
ix1=t
int1.append(ix1)
int1.append(iy1)
elif ori==2:#-
D=calc_dist(p1,p2)
tr=r1+r2
el=tr-D
a=r1-el
b=r2-el
A=a+(el/2)
B=b+(el/2)
thta=math.degrees(math.acos(A/r1))
rs=p2[1]-p1[1]
rn=p2[0]-p1[0]
gd=rs/rn
yint=p1[1]-((gd)*p1[0])
dty=calc_dist(p1,[0,yint])

aa=yint-p1[1]
bb=math.degrees(math.asin(aa/dty))
c=180-90-bb
d=180-c-thta
e=180-90-d
f=math.tan(math.radians(e))*p1[0]
g=math.sqrt(p1[0]**2+f**2)
h=g+r1
i=180-90-e
j=math.sin(math.radians(e))*h
jj=math.cos(math.radians(i))*h
k=math.cos(math.radians(e))*h
kk=math.sin(math.radians(i))*h
l=90-bb
m=90-e
tt=l+m+thta
n=(dty/math.sin(math.radians(m)))*math.sin(math.radians(thta))
nn=(g/math.sin(math.radians(l)))*math.sin(math.radians(thta))
oty=yint-n
iy1=oty+j
ix1=k
int1.append(ix1)
int1.append(iy1)

o=bb+90
p=180-o-thta
q=90-p
r=180-90-q
s=(dty/math.sin(math.radians(p)))*math.sin(math.radians(o))
t=(s/math.sin(math.radians(o)))*math.sin(math.radians(thta))
u=s+r1
v=math.sin(math.radians(r))*u
vv=math.cos(math.radians(q))*u
w=math.cos(math.radians(r))*u
ww=math.sin(math.radians(q))*u
ix2=v
otty=yint+t
iy2=otty-w
int2.append(ix2)
int2.append(iy2)

elif ori==3:#y
D=calc_dist(p1,p2)
tr=r1+r2
el=tr-D
a=r1-el
b=r2-el
A=a+(el/2)
B=b+(el/2)
b=math.sqrt(r1**2-A**2)
int1.append(p1[0]+A)
int1.append(p1[1]+b)
int2.append(p1[0]+A)
int2.append(p1[1]-b)
elif ori==4:#x
D=calc_dist(p1,p2)
tr=r1+r2
el=tr-D
a=r1-el
b=r2-el
A=a+(el/2)
B=b+(el/2)
b=math.sqrt(r1**2-A**2)
int1.append(p1[0]+b)
int1.append(p1[1]-A)
int2.append(p1[0]-b)
int2.append(p1[1]-A)
return [int1,int2]
def calc_dist(p1,p2):
return math.sqrt((p2[0] - p1[0]) ** 2 +
(p2[1] - p1[1]) ** 2)

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