Do the circles overlap?

Question

Given the coordinates of the centres and the radii of 2 circles, output a truthy value of whether they do or do not overlap.

Input

• Input may be taken via STDIN or equivalent, function arguments, but not as a variable. You can take them as a single variable (list, string etc) or as multiple inputs / arguments, in whatever order you want.

• The input will be six floats. These floats will be up to 3 decimal places. The coordinates can be positive or negative. The radii will be positive.

Output

• Output can be via STDOUT or function return.

• The program must have exactly 2 distinct outputs - one for a True value (the circles do overlap) and one for a False output (they don't overlap).

Test cases

(Input is given as list of tuples [(x1, y1, r1), (x2, y2, r2)] for the test cases; you can take input in any format)

True

[(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)]

False

[(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)]

---

This is Code Golf, shortest answer in bytes wins.

• JungHwan Min: What do we need to return if two circles are touching externally?
• dmckee --- ex-moderator kitten: The technical term for "touching but not overlapping" is "tangent" and it is a thing in geometry if nowhere else.
• Wheat Wizard: Taking floats seems like a pretty stringent requirement. Could you relax it to a more general representation? I would like to solve this in Brain-Flak, but I am unlikely to take the time to implement IEEE floats, and if I did it would be 90% of the byte count anyway so I would just be golfing a float implementation.
• Wheat Wizard: I would also like to point out that floats are not accurate up to "three decimal places" in a lot of cases. I'm not sure exactly what you want answers to handle, but its a little confusing right now.
• user45941: I think you might have a fundamental misunderstanding of how floats work. Because they are fixed-size, as the values get larger, the precision gets lower. There is a point beyond which a float cannot accurately represent all values within 3 decimal places. Also, editing a challenge to remove an unnecessary restriction is not discouraged.

Answer 1

Jelly, 5 bytes

IA<S}

Takes two complex numbers (centers) as first argument, and two real numbers (radii) as second argument.

Try it online!

How it works

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.

• hyperneutrino: Damn, I forgot about using complex numbers for coordinates. Good one! :D
• Jonathan Allan: Out of interest would the result of A here be considered the norm of the row vector "centers"? (ÆḊ itself errors with complex content.)
• Dennis: @JonathanAllan Yes, A computes the distances of the centers as the norm of their difference vector.

Answer 2

JavaScript (ES6), 38 bytes

Takes input as 6 distinct variables x1, y1, r1, x2, y2, r2.

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

Test cases

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))

• AncientSwordRage: For anyone who hasn't seem Math.hypot before.
• V. Courtois: scala polyglot :) but it does not work for some reason
• Arnauld: @V.Courtois The way you pass the parameters doesn't match the method declaration. It should be a:Double,x:Double,b:Double,y:Double,r:Double,q:Double.
• V. Courtois: @Arnauld ooh~ thanks! Should I post it separately?
• Arnauld: @V.Courtois Sure. Go for it!

Answer 3

Pyth, 5 bytes

gsE.a

Input format:

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

Try it online

How it works

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

Answer 4

MATL, 5 bytes

ZPis<

Input format is:

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

Try it online! Or verify all test cases.

How it works

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

• Dennis Jaheruddin: Not sure if it is me, but when I use your trial link and press run I get 'Error The server's response could not be decoded' -- Also Not sure if it helps, but did you think about (ab) using complex numbers like in the Jelly answer?
• Luis Mendo: @DennisJaheruddin Hey, nice to see you again here! (1) Blame caching, probably. Did you try a hard refresh? (2) I did, but I think it's also 5 bytes (-| instead of ZP)
• Dennis Jaheruddin: I suppose it is the firewall. Now i'm wondering whether an input format with something like -r2 instead of r2 would help because then you would need three differences, instead of 2 differences and an addition... I'd better run before I get drawn in too deep!
• Luis Mendo: I don't think that negating one input is acceptable as input format. If you find any problems with the Try It Online service, would you please report here?

Answer 5

R, 39 bytes

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

takes input k=c(x1,x2,y1,y2) and r=c(r1,r2); returns FALSE for tangent circles.

Try it online!

27 bytes:

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

Takes input as a matrix with the circle centers given as rows and a vector of radii.

Try it online!

• djhurio: -2 bytes function(k,r)dist(matrix(k,2))<sum(r)
• djhurio: What about dist(matrix(scan(),2))<sum(scan())?

Answer 6

Python, 40 bytes

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

Try it online!

Uses Python's complex arithmetic to compute the distance between the two centers. I'm assuming we can't take the input points directly as complex numbers, so the code expresses them like x+y*1j.

Answer 7

Python 3, 45 bytes

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

Try it online!

Answer 8

05AB1E, 6 bytes

αs--0›

Try it online!

-1 byte by using a - b > 0 rather than (reverse) b - a < 0

• Magic Octopus Urn: Oh 05AB1E does imaginary now?
• hyperneutrino: @MagicOctopusUrn Apparently /shrug

Answer 9

Python 3, 45 bytes

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

Try it online!

-8 bytes thanks to Neil/Step Hen

• micsthepick: This code works in python 2 also.
• hyperneutrino: @micsthepick Cool, thanks. It's just the way TIO does formatting.

Answer 10

APL (Dyalog), 10 bytes

Prompts for circle centers as list of two complex numbers, then for radii as list of two numbers

(+/⎕)>|-/⎕

Try it online!

(+/⎕) [is] the sum of the radii

> greater than

| the magnitude of

-/⎕ the difference in centers

Answer 11

Mathematica, 16 bytes

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

Input: [{x1, y1}, {x2, y2}, r1, r2]

---

Mathematica has a RegionIntersection builtin, but that alone is 18 bytes long...

Built-in version:

RegionIntersection@##==EmptyRegion@2&

Takes 2 Disk objects. [Disk[{x1, y1}, r1], Disk[{x2, y2}, r2]].

Answer 12

Haskell, (削除) 37 (削除ここまで) 36 bytes

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

Try it online!

Thanks @AndersKaseorg for -1 byte!

• Anders Kaseorg: Shorter as an operator: (u!v)r x y s.

Answer 13

Jelly, 12 bytes

I2+4I2¤<5S2¤

Try it online!

-2 bytes thanks to Dennis

• caird coinheringaahing: Wouldn't creating a new link with ạ/² on it same bytes?
• hyperneutrino: @cairdcoinheringaahing ?
• caird coinheringaahing: Never mind, I got 14 bytes by doing this
• Dennis: You can use I instead of reducing by absolute difference.

Answer 14

Java (OpenJDK 8), 38 bytes

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

Try it online!

Answer 15

Java 8, (削除) 41 (削除ここまで) 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.

• Olivier Grégoire: Oh! So that's why I got 3 upvotes today, but 0 when the challenge was posted? ^^ I was wondering what triggered this weird behavior ;) Since I like my answer, and you posted the same, you get a +1 as well! :p

Answer 16

Pyth, 15 bytes

<sm^-EE2 2^+EE2

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

Test suite!

Answer 17

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.

Answer 18

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.

Answer 19

C#, (削除) 50 (削除ここまで) 41 bytes

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

Saved 9 bytes thanks to @KevinCruijssen.

• Ian H.: Can't you save a few bytes there by writing (r+R)*2 instead of (r+R)+(r+R)?
• TheLethalCoder: @IanH. Yeah don't know how I missed that.
• Ian H.: Am I missing something or does this not work?
• TheLethalCoder: @IanH. I'd made a typo, the + on the RHS should have been a *.
• Ian H.: And my feedback even made that worse. Good job on the solution though!

Answer 20

Scala, 23 bytes

Thanks @Arnauld for his almost polyglot answer.

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

Try it online!

Answer 21

PostgreSQL, 41 characters

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

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ドル&&2ドル;
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

Answer 22

Java, (削除) 50 (削除ここまで) 38 bytes

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

• laszlok: Using ideas in other answers, this can be shortened to 38 like so: (x,y,r,X,Y,R)->Math.hypot(x-X,y-Y)<r+R. In fact, just realised this is the exact same as Arnauld's JavaScript answer.
• Linnea Gräf: Thanks... This answer was nevee intended to be golfed... i thought it was such a simple challenge there wouldn't be anything that can be golfed...
• Kevin Cruijssen: I'm afraid your answer is now exactly the same as the already posted answer by @OlivierGrégoire..

Answer 23

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.

Answer 24

05AB1E, 10 bytes

- ̈nOt21+θ‹

Try it online!

Answer 25

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.

Answer 26

dc, 22 bytes

9k?+_5R-d*_3R-d*+v-vzp

Try it online!

Or try the test suite!

Input is read from stdin in the format x1 x2 y1 y2 r1 r2. Note that negative numbers are written in dc with an underscore _ instead of a minus sign -.

Output is on stdout: 1 for truthy, 0 for falsey. If the two circles are tangent to one another or if they're identical, they're considered to be overlapping.

---

This requires a recent version of GNU dc which supports the R operation (rotate).

On versions of dc which don't support R, you can instead use

9k?-d*sY-d*lY+vsD+lD-vzp

(24 bytes long), with input presented on stdin in the order r1 r2 x1 x2 y1 y2. Here's a test suite for the 24-byte program, for older versions of dc

---

Explanation

9k Set precision to 9 decimal places.
? Read the input and push all 6 numbers on the stack (r2 is at the top of the stack since it's entered last).
+ Add r1 and r2
_5R Rotate the stack so that r1+r2 is now at the bottom.
-d* Compute (y1-y2)^2.
_3R Rotate the stack so that it's now: r1+r2 (y1-y2)^2 x1 x2 (top on the right)
-d* Compute (x1-x2)^2.
+ Compute (x1-x2)^2 + (y1-y2)^2.
v Take the square root to compute the distance between the centers.
- Subtract the distance between the centers from r1-r2.
v Pop one number from the stack. If it's non-negative, then compute its square root and push that back on the stack.
z Push the size of the stack onto the stack. (This is 0 if the difference was negative, and it's 1 if the difference was positive.)
p Print the item at the top of the stack.

Answer 27

Julia 0.6.0 (46 bytes)

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

Answer 28

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.

Answer 29

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?

Answer 30

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.