给定中心的坐标和2个圆的半径，请输出它们是否重叠的真实值。

输入项

• 输入可以通过STDIN或等效的函数参数获取，但不能作为变量。您可以将它们作为单个变量（列表，字符串等）或作为多个输入/参数，以任意顺序使用。

• 输入将是六个浮点数。这些浮点数最多可以保留3个小数位。坐标可以是正或负。半径将为正。

输出量

• 输出可以通过STDOUT或函数返回。

• 程序必须有2个不同的输出-一个用于True值（圆圈确实重叠），另一个用于False输出（它们不重叠）。

测试用例

（输入[(x1, y1, r1), (x2, y2, r2)]以测试用例的元组列表形式提供；您可以采用任何格式的输入）

真正

[(5.86, 3.92, 1.670), (11.8, 2.98, 4.571)]
[(8.26, -2.72, 2.488), (4.59, -2.97, 1.345)]
[(9.32, -7.77, 2.8), (6.21, -8.51, 0.4)]

假

[(4.59, -2.97, 1.345), (11.8, 2.98, 4.571)]
[(9.32, -7.77, 2.8), (4.59, -2.97, 1.345)]
[(5.86, 3.92, 1.670), (6.21, -8.51, 0.4)]

这是Code Golf，最短答案以字节为单位。

果冻，5个字节

IA<S}

将两个复数（中心）作为第一个参数，并将两个实数（半径）作为第二个参数。

在线尝试！

怎么运行的

IA<S}  Main link.
       Left argument:  [x1 + iy1, x2 + iy2]
       Right argument: [r1, r2]

I      Increments; yield (x2 - x1) + i(y2 - y1).
 A     Absolute value; yield √((x2 - x1)² + (y2 - y1)²).
   S}  Take the sum of the right argument, yielding r1 + r2.
  <    Compare the results.

JavaScript（ES6），38个字节

将输入作为6个不同的变量x1，y1，r1，x2，y2，r2。

(x,y,r,X,Y,R)=>Math.hypot(x-X,y-Y)<r+R

测试用例

let f =

(x,y,r,X,Y,R)=>Math.hypot(x-X,y-Y)<r+R

// True
console.log(f(5.86, 3.92, 1.670, 11.8, 2.98, 4.571))
console.log(f(8.26, -2.72, 2.488, 4.59, -2.97, 1.345))
console.log(f(9.32, -7.77, 2.8, 6.21, -8.51, 0.4))

// False
console.log(f(4.59, -2.97, 1.345, 11.8, 2.98, 4.571))
console.log(f(9.32, -7.77, 2.8, 4.59, -2.97, 1.345))
console.log(f(5.86, 3.92, 1.670, 6.21, -8.51, 0.4))

Pyth，5个字节

gsE.a

输入格式：

[x1, y1], [x2, y2]
r1, r2

在线尝试

怎么运行的

     Q   autoinitialized to eval(input())
   .a    L2 norm of vector difference of Q[0] and Q[1]
gsE      sum(eval(input()) >= that

MATL，5个字节

ZPis<

输入格式为：

[x1, y1]
[x2, y2]
[r1, r2]

在线尝试！或验证所有测试用例。

怎么运行的

ZP   % Take two vectors as input. Push their Euclidean distance
i    % Input the vector of radii
s    % Sum of vector
<    % Less than?

R，39个字节

function(k,r)dist(matrix(k,2,2))<sum(r)

接受输入k=c(x1,x2,y1,y2)和r=c(r1,r2); 返回FALSE正切圆。

在线尝试！

27个字节：

function(m,r)dist(m)<sum(r)

将输入作为矩阵，将圆心指定为行，并指定半径向量。

在线尝试！

Python，40个字节

lambda x,y,r,X,Y,R:abs(x-X+(y-Y)*1j)<r+R

在线尝试！

使用Python的复杂算法计算两个中心之间的距离。我假设我们不能将输入点直接当作复数，所以代码将它们表示为x+y*1j。

Python 3，45字节

lambda X,Y,R,x,y,r:(X-x)**2+(Y-y)**2<(R+r)**2

在线尝试！

05AB1E，6个字节

αs--0›

在线尝试！

-1个字节，a - b > 0而不是使用(reverse) b - a < 0

Python 3，45字节

lambda a,b,c,d,e,f:(a-d)**2+(b-e)**2<(c+f)**2

在线尝试！

-8个字节，归功于Neil / Step Hen

APL（Dyalog），10字节

提示输入圆心作为两个复数的列表，然后提示半径作为两个数的列表

(+/⎕)>|-/⎕

在线尝试！

(+/⎕) [是]半径的总和

> 比...更棒

| 的大小

-/⎕ 中心的差异

Mathematica，16个字节

Norm[#-#2]<+##3&

输入： [{x1, y1}, {x2, y2}, r1, r2]

Mathematica有一个 RegionIntersection内置功能，但仅此一项就长18个字节...

内置版本：

RegionIntersection@##==EmptyRegion@2&

带2个Disk对象。[Disk[{x1, y1}, r1], Disk[{x2, y2}, r2]]。

Haskell，37 36字节

(u#v)r x y s=(u-x)^2+(v-y)^2<(r+s)^2

在线尝试！

感谢@AndersKaseorg的-1字节！

果冻，12字节

I²+⁴I²¤<⁵S²¤

在线尝试！

-2个字节感谢Dennis

Java（OpenJDK 8），38字节

(a,b,c,x,y,z)->Math.hypot(a-x,b-y)<c+z

在线尝试！

Java 8, 41 38 bytes

(x,y,r,X,Y,R)->Math.hypot(x-X,y-Y)<r+R

Try it here.

Apparently, Java also has Math.hypot, which is 3 bytes shorter.

EDIT: Just realized this answer is now exactly the same as @OlivierGrégoire's Java 8 answer, so please upvote him instead of me if you like the 38-byte answer.

Old answer (41 bytes):

(x,y,r,X,Y,R)->(x-=X)*x+(y-=Y)*y<(r+=R)*r

Try it here.

Pyth, 15 bytes

<sm^-EE2 2^+EE2

Takes input in the order x1,x2,y1,y2,r1,r2

Test suite!

Perl 6, 13 bytes

*+*>(*-*).abs

Try it online!

The first two arguments are the radii, in either order. The third and fourth arguments are the coordinates of the centers, as complex numbers, in either order.

Taxi, 1582 bytes

Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to The Babelfishery.Pickup a passenger going to Tom's Trims.Pickup a passenger going to Tom's Trims.Go to Tom's Trims:n.[a]Go to Post Office:s.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 l 1 r.Pickup a passenger going to What's The Difference.Pickup a passenger going to What's The Difference.Go to What's The Difference:n 5 l.Pickup a passenger going to Cyclone.Go to Cyclone:e 1 r.Pickup a passenger going to Multiplication Station.Pickup a passenger going to Multiplication Station.Go to Multiplication Station:s 1 l 2 r 4 l.Pickup a passenger going to Addition Alley.Go to Tom's Trims:s 1 r 3 r.Pickup a passenger going to The Babelfishery.Switch to plan "b" if no one is waiting.Switch to plan "a".[b]Go to Addition Alley:n 1 r 1 l 3 l 1 l.Pickup a passenger going to Magic Eight.Go to Post Office:n 1 r 1 r 3 r 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 l 1 r.Pickup a passenger going to Addition Alley.Pickup a passenger going to Addition Alley.Go to Addition Alley:n 5 l 1 l.Pickup a passenger going to Cyclone.Go to Cyclone:n 1 l 1 l.Pickup a passenger going to Multiplication Station.Pickup a passenger going to Multiplication Station.Go to Multiplication Station:s 1 l 2 r 4 l.Pickup a passenger going to Magic Eight.Go to Magic Eight:s 1 r.Switch to plan "c" if no one is waiting.'1' is waiting at Writer's Depot.[c]'0' is waiting at Writer's Depot.Go to Writer's Depot:w 1 l 2 l.Pickup a passenger going to Post Office.Go to Post Office:n 1 r 2 r 1 l.

Try it online!

Outputs 1 for overlapping circles.
Outputs 0 for non-overlapping circles (including tangential circles).

Ungolfed / formatted:

Go to Post Office: west 1st left 1st right 1st left.
Pickup a passenger going to The Babelfishery.
Pickup a passenger going to Tom's Trims.
Pickup a passenger going to Tom's Trims.
Go to Tom's Trims: north.
[a]
Go to Post Office: south.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery: south 1st left 1st right.
Pickup a passenger going to What's The Difference.
Pickup a passenger going to What's The Difference.
Go to What's The Difference: north 5th left.
Pickup a passenger going to Cyclone.
Go to Cyclone: east 1st right.
Pickup a passenger going to Multiplication Station.
Pickup a passenger going to Multiplication Station.
Go to Multiplication Station: south 1st left 2nd right 4th left.
Pickup a passenger going to Addition Alley.
Go to Tom's Trims: south 1st right 3rd right.
Pickup a passenger going to The Babelfishery.
Switch to plan "b" if no one is waiting.
Switch to plan "a".
[b]
Go to Addition Alley: north 1st right 1st left 3rd left 1st left.
Pickup a passenger going to Magic Eight.
Go to Post Office: north 1st right 1st right 3rd right 1st left.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery: south 1st left 1st right.
Pickup a passenger going to Addition Alley.
Pickup a passenger going to Addition Alley.
Go to Addition Alley: north 5th left 1st left.
Pickup a passenger going to Cyclone.
Go to Cyclone: north 1st left 1st left.
Pickup a passenger going to Multiplication Station.
Pickup a passenger going to Multiplication Station.
Go to Multiplication Station: south 1st left 2nd right 4th left.
Pickup a passenger going to Magic Eight.
Go to Magic Eight: south 1st right.
Switch to plan "c" if no one is waiting.
'1' is waiting at Writer's Depot.
[c]
'0' is waiting at Writer's Depot.
Go to Writer's Depot: west 1st left 2nd left.
Pickup a passenger going to Post Office.
Go to Post Office: north 1st right 2nd right 1st left.

C#, 50 41 bytes

(x,y,r,X,Y,R)=>(x-=X)*x+(y-=Y)*y<(r+=R)*r

Saved 9 bytes thanks to @KevinCruijssen.

Scala, 23 bytes

Thanks @Arnauld for his almost polyglot answer.

math.hypot(a-x,b-y)<r+q

Try it online!

PostgreSQL, 41 characters

prepare f(circle,circle)as select $1&&$2;

Prepared statement, takes input as 2 parameters in any circle notation.

Sample run:

Tuples only is on.
Output format is unaligned.
psql (9.6.3, server 9.4.8)
Type "help" for help.

psql=# prepare f(circle,circle)as select $1&&$2;
PREPARE

psql=# execute f('5.86, 3.92, 1.670', '11.8, 2.98, 4.571');
t

psql=# execute f('8.26, -2.72, 2.488', '4.59, -2.97, 1.345');
t

psql=# execute f('9.32, -7.77, 2.8', '6.21, -8.51, 0.4');
t

psql=# execute f('4.59, -2.97, 1.345', '11.8, 2.98, 4.571');
f

psql=# execute f('9.32, -7.77, 2.8', '4.59, -2.97, 1.345');
f

psql=# execute f('5.86, 3.92, 1.670', '6.21, -8.51, 0.4');
f

Java, 50 38 bytes

(x,y,r,X,Y,R)->Math.hypot(x-X,y-Y)<r+R

x86 Machine Code (with SSE2), 36 bytes

; bool CirclesOverlap(double x1, double y1, double r1,
;                     double x2, double y2, double r2);
F2 0F 5C C3        subsd   xmm0, xmm3      ; x1 - x2
F2 0F 5C CC        subsd   xmm1, xmm4      ; y1 - y2
F2 0F 58 D5        addsd   xmm2, xmm5      ; r1 + r2
F2 0F 59 C0        mulsd   xmm0, xmm0      ; (x1 - x2)^2
F2 0F 59 C9        mulsd   xmm1, xmm1      ; (y1 - y2)^2
F2 0F 59 D2        mulsd   xmm2, xmm2      ; (r1 + r2)^2
F2 0F 58 C1        addsd   xmm0, xmm1      ; (x1 - x2)^2 + (y1 - y2)^2
66 0F 2F D0        comisd  xmm2, xmm0
0F 97 C0           seta    al              ; ((r1 + r2)^2) > ((x1 - x2)^2 + (y1 - y2)^2)
C3                 ret

The above function accepts descriptions of two circles (x- and y-coordinates of center point and a radius), and returns a Boolean value indicating whether or not they intersect.

It uses a vector calling convention, where the parameters are passed in SIMD registers. On x86-32 and 64-bit Windows, this is the __vectorcall calling convention. On 64-bit Unix/Linux/Gnu, this is the standard System V AMD64 calling convention.

The return value is left in the low byte of EAX, as is standard with all x86 calling conventions.

This code works equally well on 32-bit and 64-bit x86 processors, as long as they support the SSE2 instruction set (which would be Intel Pentium 4 and later, or AMD Athlon 64 and later).

AVX version, still 36 bytes

If you were targeting AVX, you would probably want to add a VEX prefix to the instructions. This does not change the byte count; just the actual bytes used to encode the instructions:

; bool CirclesOverlap(double x1, double y1, double r1,
;                     double x2, double y2, double r2);
C5 FB 5C C3      vsubsd   xmm0, xmm0, xmm3   ; x1 - x2
C5 F3 5C CC      vsubsd   xmm1, xmm1, xmm4   ; y1 - y2
C5 EB 58 D5      vaddsd   xmm2, xmm2, xmm5   ; r1 + r2
C5 FB 59 C0      vmulsd   xmm0, xmm0, xmm0   ; (x1 - x2)^2
C5 F3 59 C9      vmulsd   xmm1, xmm1, xmm1   ; (y1 - y2)^2
C5 EB 59 D2      vmulsd   xmm2, xmm2, xmm2   ; (r1 + r2)^2
C5 FB 58 C1      vaddsd   xmm0, xmm0, xmm1   ; (x1 - x2)^2 + (y1 - y2)^2
C5 F9 2F D0      vcomisd  xmm2, xmm0
0F 97 C0         seta     al                 ; ((r1 + r2)^2) > ((x1 - x2)^2 + (y1 - y2)^2)
C3               ret

AVX instructions have the advantage of taking three operands, allowing you to do non-destructive operations, but that doesn't really help us to compact the code any here. However, mixing instructions with and without VEX prefixes can result in sub-optimal code, so you generally want to stick with all AVX instructions if you're targeting AVX, and in this case, it doesn't even hurt your byte count.

05AB1E, 10 bytes

-¨nOt²¹+θ‹

Try it online!

PHP, 66 bytes

<?php $i=$argv;echo hypot($i[1]-$i[4],$i[2]-$i[5])<$i[3]+$i[6]?:0;

Try it online!

Runs from the command line, taking input as 6 command-line parameter arguments, and prints 1 if the circles overlap, else 0.

Julia 0.6.0 (46 bytes)

a->((a[1]-a[2])^2+(a[3]-a[4])^2<(a[5]+a[6])^2)

Clojure, 68 bytes

#(<(+(*(- %4 %)(- %4 %))(*(- %5 %2)(- %5 %2)))(*(+ %6 %3)(+ %6 %3)))

Takes six arguments: x1, y1, r1, x2, y2, r2. Returns true or false.

Sadly, Clojure does not have a pow function of some sorts. Costs a lot of bytes.

Actually, 8 bytes

-)-(h@+>

Try it online!

Explanation:

-)-(h@+>  (implicit input: [y1, y2, x1, x2, r1, r2])
-         y2-y1 ([y2-y1, x1, x2, r1, r2])
 )-       move to bottom, x1-x2 ([x1-x2, r1, r2, y2-y1])
   (h     move from bottom, Euclidean norm ([sqrt((y2-y1)**2+(x2-x1)**2), r1, r2])
     @+   r1+r2 ([r1+r2, norm])
       >  is r1+r2 greater than norm?

R (+pryr), 31 bytes

pryr::f(sum((x-y)^2)^.5<sum(r))

Which evaluates to the function

function (x, y, z) 
sum((x - y)^2)^0.5 < sum(z)

Where x are the coordinates of circle 1, y are the coordinates of circle 2 and z the radii.

Calculates the distance between the two centers using Pythagoras and tests if that distance is smaller than the sum of the radii.

Makes use of R's vectorisation to simultaneously calculate (x1-x2)^2 and (y1-y2)^2. These are then summed and squarely rooted.

Go, 93 bytes

package q
import c "math/cmplx"
func o(a,b complex128,r,R float64)bool{return c.Abs(b-a)<r+R}

Fairly simple algorithm, same as several other answers, except it uses the built-in complex type and calls math/cmplx.Abs().

Taking the radii as complex numbers doesn't help, because the cast to float64 adds more bytes than the variable declaration saves (can't do float64 < complex128).

Try it online! Includes the test cases, and uses package main instead of a library.