How to find out if two circles intersect each other

Question

I want to find the intersecting of 2 circles with radius r1 and r2 (simple solutions preferred).

Answer 1

Tetrad covered general intersection in his post. Here I'll cover an algorithm that returns the specific points of intersection based on the formulae in this concise article. I'm matching my variable names to those in the article, so keep this diagram in mind - and probably in view too!

The language is Python. You can verify your results in Wolfram Alpha by running a query to determine the intersection of two circles like this:

intersection ((x - 0)^2 + (y - 0)^2 = 2^2), ((x - 1)^2 + (y - 0)^2 = 2^2)

or in the general case

intersection ((x - h)^2 + (y - k)^2 = r^2), ((x - h)^2 + (y - k)^2 = r^2)

where for each of the two circles, h = the x-coordinate of the centre of the circle, k = the y-coordinate, and r is the radius.

from math import sqrt

# Determines whether two circles collide and, if applicable,
# the points at which their borders intersect.
# Based on an algorithm described by Paul Bourke:
# http://local.wasp.uwa.edu.au/~pbourke/geometry/2circle/
# Arguments:
#   P0 (complex): the centre point of the first circle
#   P1 (complex): the centre point of the second circle
#   r0 (numeric): radius of the first circle
#   r1 (numeric): radius of the second circle
# Returns:
#   False if the circles do not collide
#   True if one circle wholly contains another such that the borders
#       do not overlap, or overlap exactly (e.g. two identical circles)
#   An array of two complex numbers containing the intersection points
#       if the circle's borders intersect.
def IntersectPoints(P0, P1, r0, r1):
    if type(P0) != complex or type(P1) != complex:
        raise TypeError("P0 and P1 must be complex types")
    # d = distance
    d = sqrt((P1.real - P0.real)**2 + (P1.imag - P0.imag)**2)
    # n**2 in Python means "n to the power of 2"
    # note: d = a + b

    if d > (r0 + r1):
        return False
    elif d < abs(r0 - r1):
        return True
    elif d == 0:
        return True
    else:
        a = (r0**2 - r1**2 + d**2) / (2 * d)
        b = d - a
        h = sqrt(r0**2 - a**2)
        P2 = P0 + a * (P1 - P0) / d

        i1x = P2.real + h * (P1.imag - P0.imag) / d
        i1y = P2.imag - h * (P1.real - P0.real) / d
        i2x = P2.real - h * (P1.imag - P0.imag) / d
        i2y = P2.imag + h * (P1.real - P0.real) / d

        i1 = complex(i1x, i1y)
        i2 = complex(i2x, i2y)

        return [i1, i2]

def CompToStr(c):
    return "(" + str(c.real) + ", " + str(c.imag) + ")"

def PairToStr(p):
    return CompToStr(p[0]) + " , " + CompToStr(p[1])

def Test():
    ip = IntersectPoints

    i = ip(complex(0,0), complex(1, 0), 2, 2)
    s = ip(complex(0,0), complex(4, 0), 2, 2)

    print "Intersection:", PairToStr(i)
    print "Wholly inside:", ip(complex(0,0), complex(1, 0), 5, 2)
    print "Single-point edge collision:", PairToStr(s)
    print "No collision:", ip(complex(0,0), complex(5, 0), 2, 2)

Test()

Note that this algorithm uses the readily-available complex class for Python which can mimic a 2D Vector's x and y via the complex.real and complex.imag attributes respectively. Mathematically speaking a complex is a 2D vector on the complex plane, which is why this works, so programmatically only the syntax is different. It's not optimal but I don't think Python natively has a proper geometric vector class.

Whether this was useful to you or not it was still fun. ;)

Answer 2

Do you mean if the circles are overlapping? That's easy, just figure out the distance between the two and compare it to the r1 and r2. Or more specifically, calculate the squared distance between the two (it's faster) and compare it to (r1 + r2)^2.

If you mean what the actual intersection points are of the two circles, then I don't know the math for that offhand.

Answer 3

Do you really need the actual intersection area or do you just need to know if you're in the intersection area? The latter is easier to solve--if you're closer than the radius to both centers you're in the intersection area.