Do the circles intersect?

Question

This question is similar to Do the circles overlap?, except your program should return True if the circumference of one circle intersects the other, and false if they do not. If they touch, this is considered an intersection. Note that concentric circles only intersect when both radii are equal

Input is two x, y coordinates for the circle centres and two radius lengths, as either 6 floats or ints, in any order, and output a boolean value, (you may print True or False if you choose)

test cases are in format (x1, y1 x2, y2, r1, r2):

these inputs should output true:

0 0 2 2 2 2
1 1 1 5 2 2
0.55 0.57 1.97 2.24 1.97 2.12

these should output false:

0 0 0 1 5 3
0 0 5 5 2 2
1.57 3.29 4.45 6.67 1.09 3.25

Answer 1

APL (Dyalog), 20 bytes

Uses Anders Kaseorg's formula: 0≤(x₁​−x₂)²​−(r₁​−r₂)²+(y₁​−y₂)²≤4​r₁r₂

Takes (x₁, r₁, y₁) as left argument and (x₂, r₂, y₂) as right argument.

(-/2*⍨-)(≤∧0≤⊣)4×2⊃×

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The overall function's structure is a fork (3-train) where the tines are -/2*⍨- and ≤∧0≤⊣ and 4×2⊃×*. The middle tine takes the results of the side tines as arguments. The side tines use the overall function's arguments.

Right tine:
× multiply the arguments (x₁x₂, r₁r₂, y₁y₂)
2⊃ pick the second element (r₁r₂)
4× multiply by four (4​r₁r₂)

Left tine:
- subtract the arguments (x₁​−x₂, r₁​−r₂, y₁​−y₂)
2*⍨ square ((x₁​−x₂)², (r₁​−r₂)², (y₁​−y₂)²)
-/ minus reduction i.e. alternate sum* ((x₁​−x₂)²​−((r₁​−r₂)²​−(y₁​−y₂)²))

Now we use these results as arguments to the middle tine:
0≤⊣ is the left argument greater than or equal to zero
∧ and
≤ the left argument smaller than or equal to the right argument?

---

* due to APL's right associativity

Answer 2

Mathematica 67 bytes

This checks whether the area of the intersection of two disks is positive.

RegionMeasure@RegionIntersection[{#,#2}~Disk~#3,{#4,#5}~Disk~#6]>0&

Answer 3

APL (Dyalog Classic), 17 bytes

(+/∧.≥+⍨)⎕,|0j1⊥⎕

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expects two lines of input: x1 y1 x2 y2 and r1 r2

⎕ read and evaluate a line

0j1 imaginary constant i=sqrt(-1)

0j1⊥ decode from base-i, thus computing: i³x₁ + i²y₁ + i¹x₂ + i⁰y₂ = -i x₁ - y₁ + ix₂ + y₂ = i(x₂-x₁) + (y₂-y₁)

| magnitude of that complex number, i.e. distance between the two points

⎕, read the two radii and prepend them, so we have a list of 3 reals

(+/∧.≥+⍨) flat-tolerant triangle inequality: the sum (+/) must be greater than or equal to (≥) the double of each side (+⍨), and all these results must be true (∧.)

Answer 4

Python 3, 56 55 bytes

lambda X,Y,x,y,R,r:0<=(X-x)**2+(Y-y)**2-(R-r)**2<=4*r*R

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Answer 5

GolfScript, 20 bytes

~@- 2?@@- 2?+@@+2?>!

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Takes arguments in the following form r1 r2 x1 y1 x2 y2

Implements the formula below

GolfScript Does not inherently support floating point types. See this post for more details about passing floats https://codegolf.stackexchange.com/a/26553

My original intention was to use GolfScript's zip and fold operations.

GolfScript, 26 bytes

1:a;~zip{{a*+2?-1:a;}*}/+>

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Answer 6

JavaScript, 52 bytes

(X,Y,x,y,R,r)=>4*r*R/((X-x)**2+(Y-y)**2-(R-r)**2)>=1

Some how Based on Anders Kaseorg Python solution

Answer 7

Scala, 59 bytes

This might be golfable.

val u=(a-x)*(a-x)+(b-y)*(b-y)
(q-r)*(q-r)<=u&u<=(q+r)*(q+r)

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Scala, 66 bytes

I do like this one, but sadly the previous was shorter. I hate Scala for forcing defining parameter type...

var p=math.pow(_:Double,2)
val u=p(a-x)+p(b-y)
p(q-r)<=u&u<=p(q+r)

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