Computational Geometry contents

Common Tangents to Two Circles

Find all lines touching two circles at once (up to four), or the tangents from a point to a circle, with a small algebraic derivation.

Intermediate6 min readgeometrycirclestangentlines

Read first: Equation of a Line Through a Segment

Given two circles, find all lines that touch both: the common tangents. The number of common tangents is 4, 3, 2, 1, 0 or infinite depending on the arrangement:

Arrangement	Common tangents
disjoint, outside each other	4 (two outer, two inner)
externally tangent	3 (one is the shared tangent, counted twice)
overlapping in two points	2
internally tangent	1
one strictly inside the other	0
identical circles	infinitely many

The algorithm below computes up to four candidate lines. In degenerate arrangements some candidates coincide (the same line is produced twice, possibly with opposite signs) or do not exist, so we deduplicate at the end; the infinite case (identical circles) must be handled separately. It also works if one or both radii are $0$ : a circle of radius $0$ is a point, so this gives the two tangents from a point to a circle, or the single line through two points.

Algebraic derivation

Translate so the first circle is centered at the origin. Let $r_1, r_2$ be the radii and $v = (v_x, v_y) \ne 0$ the center of the second circle. We look for lines $ax + by + c = 0$ with $a^2 + b^2 = 1$ (normalized, so that $ax + by + c$ is the signed distance to the line) at distance $r_1$ from the origin and distance $r_2$ from $v$ :

a^2 + b^2 = 1,\qquad |c| = r_1,\qquad |a v_x + b v_y + c| = r_2

Opening the absolute values gives sign choices: $c = d_1 = \pm r_1$ , $a v_x + b v_y + c = d_2 = \pm r_2$ . It is a quadratic system whose solutions are

a = \frac{(d_2 - d_1)v_x \pm v_y\sqrt{v_x^2 + v_y^2 - (d_2-d_1)^2}}{v_x^2 + v_y^2},\quad b = \frac{(d_2 - d_1)v_y \mp v_x\sqrt{v_x^2 + v_y^2 - (d_2-d_1)^2}}{v_x^2 + v_y^2},\quad c = d_1

Flipping the sign of $(a, b, c)$ describes the same line with the two circles on the other side, which is why only one of the two square-root signs is needed for each of the four $(d_1, d_2)$ combinations. Finally, if the first circle was at $(x_0, y_0)$ , subtract $a x_0 + b y_0$ from $c$ .

Implementation

python

import math

EPS = 1e-9

def tangent_candidates(c1, r1, c2, r2):
    """Up to four normalized lines (a, b, c), a*x + b*y + c = 0 with a^2 + b^2 = 1."""
    vx, vy = c2[0] - c1[0], c2[1] - c1[1]
    z = vx * vx + vy * vy
    lines = []
    for s1 in (-1, 1):
        for s2 in (-1, 1):
            d1, d2 = s1 * r1, s2 * r2
            r = d2 - d1
            d = z - r * r
            if d < -EPS:
                continue                       # no line for this choice of sides
            d = math.sqrt(abs(d))
            a = (vx * r + vy * d) / z
            b = (vy * r - vx * d) / z
            c = d1 - (a * c1[0] + b * c1[1])   # undo the translation of the first center
            lines.append((a, b, c))
    return lines

def dist_to_line(line, p):
    return abs(line[0] * p[0] + line[1] * p[1] + line[2])

# far apart: four different lines, each at distance r1 from the first center and r2 from the second
lines = tangent_candidates((0, 0), 1, (10, 0), 1)
assert len(lines) == 4
for l in lines:
    assert abs(l[0] ** 2 + l[1] ** 2 - 1) < 1e-9
    assert abs(dist_to_line(l, (0, 0)) - 1) < 1e-9 and abs(dist_to_line(l, (10, 0)) - 1) < 1e-9

The lines (a, b, c) and (-a, -b, -c) are the same line. To compare or count lines, put them in a canonical form (a fixed sign) and remove duplicates:

python

def canonical(line):
    a, b, c = line
    if a < -EPS or (abs(a) <= EPS and b < 0):
        a, b, c = -a, -b, -c
    return (round(a, 7) + 0.0, round(b, 7) + 0.0, round(c, 7) + 0.0)

def common_tangents(c1, r1, c2, r2):
    """Distinct common tangents of two different circles (or points, if a radius is 0)."""
    return sorted({canonical(l) for l in tangent_candidates(c1, r1, c2, r2)})

def check(c1, r1, c2, r2, expected):
    lines = common_tangents(c1, r1, c2, r2)
    assert len(lines) == expected, (len(lines), expected)
    for l in lines:
        assert abs(dist_to_line(l, c1) - r1) < 1e-6 and abs(dist_to_line(l, c2) - r2) < 1e-6
    return lines

check((0, 0), 1, (10, 0), 1, 4)          # disjoint: 2 outer + 2 inner
check((0, 0), 2, (10, 3), 1, 4)
check((0, 0), 1, (2, 0), 1, 3)           # externally tangent: the two inner tangents coincide
check((0, 0), 3, (4, 0), 3, 2)           # overlapping: only the two outer tangents
check((0, 0), 3, (1, 0), 2, 1)           # internally tangent: a single tangent, at the touching point
check((0, 0), 5, (1, 0), 1, 0)           # one circle inside the other: none

Tangents from a point

Set one radius to $0$ . A point outside a circle has two tangent lines, on the circle the tangent is unique, and inside there are none:

python

check((5, 0), 0, (0, 0), 3, 2)           # point (5, 0) outside the circle of radius 3
check((3, 0), 0, (0, 0), 3, 1)           # on the circle
check((1, 0), 0, (0, 0), 3, 0)           # inside the circle
check((0, 0), 0, (5, 5), 0, 1)           # two points: the line through them

The touching points are easy to get from the line: the foot of the perpendicular from the center, i.e. the center minus the signed distance times the normal.

python

def touching_point(line, center):
    a, b, c = line
    s = a * center[0] + b * center[1] + c            # signed distance to the line
    return (center[0] - s * a, center[1] - s * b)

for l in common_tangents((0, 0), 2, (10, 3), 1):
    p, q = touching_point(l, (0, 0)), touching_point(l, (10, 3))
    assert abs(math.dist(p, (0, 0)) - 2) < 1e-6 and abs(math.dist(q, (10, 3)) - 1) < 1e-6
    assert dist_to_line(l, p) < 1e-6 and dist_to_line(l, q) < 1e-6

Randomized check

Every returned line must be at exactly the right distance from both centers, and the number of distinct tangents must match the classification by the distance $d$ between the centers (for $d>0$ ):

Condition	Tangents
$d > r_1 + r_2$	4
$d = r_1 + r_2$	3
$\lvert r_1 - r_2\rvert < d < r_1 + r_2$	2
$d = \lvert r_1 - r_2\rvert$	1
$d < \lvert r_1 - r_2\rvert$	0

python

import random

rnd = random.Random(8)
for _ in range(3000):
    c1 = (rnd.randint(-8, 8), rnd.randint(-8, 8))
    c2 = (rnd.randint(-8, 8), rnd.randint(-8, 8))
    r1, r2 = rnd.randint(1, 6), rnd.randint(1, 6)
    d2 = (c1[0] - c2[0]) ** 2 + (c1[1] - c2[1]) ** 2
    if d2 == 0:
        continue                                                # concentric circles: handled separately
    if d2 > (r1 + r2) ** 2:
        expected = 4
    elif d2 == (r1 + r2) ** 2:
        expected = 3
    elif d2 > (r1 - r2) ** 2:
        expected = 2
    elif d2 == (r1 - r2) ** 2:
        expected = 1
    else:
        expected = 0
    check(c1, r1, c2, r2, expected)