Circle Tangents — Mathematical Foundations¶

One-line summary: A line ℓ: a·x + b·y + c = 0 with a²+b²=1 is tangent to circle (Cᵢ, rᵢ) iff a·Cᵢx + b·Cᵢy + c = σᵢrᵢ for some σᵢ ∈ {±1}. Solving the four sign-pairs gives a quadratic whose discriminant is d² − (σ1r1 − σ2r2)²; its sign reproduces exactly the 0..4 common-tangent count, which we prove rigorously and connect to the homothety (external/internal similarity-center) structure and to degeneracy via exact predicates.

Table of Contents¶

1. Formal Definitions
2. Tangent from an External Point — Derivation
3. The Common-Tangent System and Its Solution
4. The 0–4 Count — Theorem and Proof
5. The Homothety / Similarity-Center Structure
6. Touch-Point Formulas and Their Correctness
7. Degeneracy and Exact Predicates
8. Closed-Form Lengths and Angles of Common Tangents
9. A Fully Worked Numeric Derivation
10. Parametrization and the Touch-Point Locus
11. Error Bounds and Conditioning
12. Tangent Lines and the Apollonius / Power-of-a-Point Structure
13. The Count as a Morse-Theoretic Bifurcation
14. A Rigorous Proof that Radius ⟂ Tangent
15. Generalized Belt: Convex Hull of Discs
16. Trigonometric Closed Forms for Touch-Point Angles
17. Oriented (Signed) Tangents for Path Planning
18. Comparison with Alternatives
19. Summary

Formal Definitions¶

Circle:   ◯(C, r) = { X ∈ ℝ² : |X − C| = r },  r > 0,  C ∈ ℝ².
Disc:     D(C, r) = { X ∈ ℝ² : |X − C| ≤ r }.

Line in normal form:
   ℓ(a, b, c) = { X = (x, y) : a·x + b·y + c = 0 },  with  a² + b² = 1.
   The vector n = (a, b) is the UNIT NORMAL; |n| = 1.

Signed distance from a point Q to ℓ:
   sd(Q, ℓ) = a·Qx + b·Qy + c.
   |sd(Q, ℓ)| = Euclidean distance from Q to ℓ;
   sign(sd) selects the half-plane containing Q.

Tangency:  ℓ is tangent to ◯(C, r)  ⟺  |sd(C, ℓ)| = r
           ⟺  sd(C, ℓ) = σ·r  for a unique σ ∈ {+1, −1}.
           The TOUCH-POINT is  T = C − sd(C, ℓ)·n,  and |T − C| = r, T ∈ ℓ.

External point P to ◯(C, r):  |P − C| = d > r.
External common tangent of two circles:  σ1 = σ2 (centers same side).
Internal common tangent:                 σ1 = −σ2 (centers opposite sides).

Lemma (closest-point characterization of tangency). A line ℓ is tangent to ◯(C, r) iff the foot of the perpendicular from C to ℓ lies on the circle, i.e. dist(C, ℓ) = r.

Proof. dist(C, ℓ) = min_{X∈ℓ} |X − C|, attained at the perpendicular foot F. If dist(C,ℓ) = r, then F is on the circle and every other point of ℓ is at distance > r, hence outside the circle — so ℓ ∩ ◯ = {F}, one point: tangent. Conversely if ℓ is tangent, the unique common point is the closest point of ℓ to C (a strictly closer point would be inside the circle, forcing a second crossing), so dist(C,ℓ) = r. ∎

This Lemma is the bedrock: it converts "tangent" into the algebraic equation sd(C,ℓ) = ±r, and it proves radius ⟂ tangent (C → F is perpendicular to ℓ by definition of perpendicular foot).

Tangent from an External Point¶

Theorem 1. Let P be external to ◯(C, r), d = |P − C| > r. Then there are exactly two tangent lines from P, their touch-points T satisfy |P − T| = sqrt(d² − r²), and the angle α = ∠TPC satisfies sin α = r/d.

Proof. By the Lemma, a tangent touches at T with CT ⟂ PT, so triangle △PTC is right-angled at T. The hypotenuse is PC = d, the leg CT = r. By the Pythagorean theorem |PT|² = d² − r², giving the tangent length L = sqrt(d² − r²) (real since d > r). The angle at P opposite the leg CT satisfies sin α = opposite/hypotenuse = r/d.

Existence of exactly two: the locus of points T with ∠PTC = 90° is the circle with diameter PC (Thales' theorem), call it Θ with center M = (P+C)/2, radius d/2. The touch-points are ◯(C,r) ∩ Θ. Two distinct circles intersect in 0, 1, or 2 points; here |M − C| = d/2, Θ has radius d/2, ◯ has radius r. The intersection count condition |R−ρ| < |M−C| < R+ρ becomes |d/2 − r| < d/2 < d/2 + r, i.e. −r < d/2 − d/2 ... — concretely |d/2 − r| < d/2 ⟺ 0 < r < d, which holds since 0 < r < d. Hence exactly two intersection points: two tangents. ∎

Touch-point closed form. With φ = atan2(P_y − C_y, P_x − C_x) (direction C→P) and θ = acos(r/d):

T± = C + r·(cos(φ ± θ), sin(φ ± θ)).

Why θ = acos(r/d): in △PTC the angle at the center C (i.e. ∠PCT) has adjacent leg CT = r along… more precisely, project: cos(∠PCT) = (CT)/(CP) · .... Cleanly, cos θ = r/d because the right angle is at T: cos(∠PCT) = adjacent/hypotenuse = CT/CP = r/d. The two touch-points are obtained by rotating the direction C→P by ±θ and scaling to radius r. The degenerate d = r gives θ = 0: the two points coincide at the single touch-point, consistent with P lying on the circle. The case d < r gives r/d > 1, acos undefined — no real tangent, matching P interior. ∎

The Common-Tangent System¶

For two circles ◯(C₁, r₁), ◯(C₂, r₂), a common tangent ℓ(a,b,c) (with a²+b²=1) must satisfy, for some σ₁, σ₂ ∈ {±1}:

(1)   a·C₁x + b·C₁y + c = σ₁ r₁
(2)   a·C₂x + b·C₂y + c = σ₂ r₂
(3)   a² + b² = 1

Subtract (1) from (2) to eliminate c. Let Δ = C₂ − C₁ = (Δx, Δy), d = |Δ|, and define

g = σ₂ r₂ − σ₁ r₁.        (the radius gap for this sign-pair)

Then

(4)   a·Δx + b·Δy = g,     with    a² + b² = 1.

Equation (4) says the unit vector n = (a,b) has prescribed projection g/d onto the unit direction Δ/d. Decompose n into components along û = Δ/d and along û⊥ = (−Δy, Δx)/d:

n = (g/d)·û  +  h·û⊥,     where  h = ±sqrt(1 − (g/d)²).

h is real iff (g/d)² ≤ 1, i.e. |g| ≤ d. Writing it out:

a = (g·Δx − sign·h·d·Δy)/d²  ... equivalently
a = Δx·g/d² ∓ Δy·sqrt(d²−g²)/d²
b = Δy·g/d² ± Δx·sqrt(d²−g²)/d²

and c = σ₁ r₁ − a·C₁x − b·C₁y from (1). This is exactly the middle.md solver. The number of real n per sign-pair is:

|g| < d   → 2 solutions  (h ≠ 0, both ±)
|g| = d   → 1 solution   (h = 0, the two coincide)
|g| > d   → 0 solutions  (h imaginary)

Theorem 2 (existence test per type). - External (σ₁ = σ₂): g = ±(r₂ − r₁), |g| = |r₁ − r₂|. External tangents exist iff |r₁ − r₂| ≤ d. - Internal (σ₁ = −σ₂): g = ±(r₁ + r₂), |g| = r₁ + r₂. Internal tangents exist iff r₁ + r₂ ≤ d.

Each ±g of the same magnitude yields the same set of lines (it merely relabels which side), so a type contributes two lines when its |g| < d strictly, one when |g| = d, zero when |g| > d. ∎

The 0–4 Count Theorem¶

Theorem 3. Let d = |C₂ − C₁|, and WLOG r₁ ≥ r₂ > 0. The number of common tangent lines of ◯(C₁,r₁) and ◯(C₂,r₂) is:

(Concentric d = 0 with r₁ ≠ r₂ is the sub-case of "nested" → 0; identical circles d=0, r₁=r₂ have infinitely many and are excluded.)

Proof. Apply Theorem 2 to each type.

External tangents (|g| = r₁ − r₂, using r₁ ≥ r₂ so |g| = r₁ − r₂): - d < r₁ − r₂: |g| > d → 0 external. - d = r₁ − r₂: |g| = d → 1 external (the unique tangent at the internal contact point). - d > r₁ − r₂: |g| < d → 2 external.

Internal tangents (|g| = r₁ + r₂): - d < r₁ + r₂: |g| > d → 0 internal. - d = r₁ + r₂: |g| = d → 1 internal (the unique tangent at the external contact point). - d > r₁ + r₂: |g| < d → 2 internal.

Now combine across the five d-regimes (note r₁ − r₂ ≤ r₁ + r₂ always, with equality iff r₂ = 0):

d < r₁−r₂          : ext 0, int 0 → 0
d = r₁−r₂          : ext 1, int 0 → 1
r₁−r₂ < d < r₁+r₂  : ext 2, int 0 → 2
d = r₁+r₂          : ext 2, int 1 → 3
d > r₁+r₂          : ext 2, int 1 → 4

These are precisely the table rows. The transitions are continuous in d: as d increases through r₁−r₂, two external tangents are born from one (the discriminant d²−(r₁−r₂)² crosses zero); as d increases through r₁+r₂, two internal tangents are born from one. ∎

Corollary (duality with circle–circle intersection). Let k(d) = # intersection points of the two circles and t(d) = # common tangents. Then t(d) + (jumps) track inversely: where k goes 0,1,2,1,0 over the regimes nested→tangent→overlap→tangent→separate, t goes 0,1,2,3,4. Both are governed by the same two thresholds r₁±r₂, because both reduce to the sign of d²−(r₁−r₂)² and d²−(r₁+r₂)². The intersection point count solves a²+b²=1 with sd = +r both circles crossing (chord), whereas the tangent count solves sd = σr; the shared discriminants are why the classification tables coincide.

The Homothety Structure¶

A homothety H(O, k) with center O and ratio k ≠ 0 maps X ↦ O + k(X − O). It maps ◯(C, r) to ◯(O + k(C−O), |k|r).

Proposition (similarity centers). For ◯(C₁,r₁), ◯(C₂,r₂) with r₁ ≠ r₂: - The external center E = (r₂C₁ − r₁C₂)/(r₂ − r₁) is the center of the homothety with ratio k = r₂/r₁ > 0 mapping circle 1 to circle 2. - The internal center I = (r₂C₁ + r₁C₂)/(r₁ + r₂) is the center of the homothety with ratio k = −r₂/r₁ < 0.

Proof sketch. E is the unique point with (E − C₂) = (r₂/r₁)(E − C₁) (collinear, external division in ratio r₁:r₂); solving gives the stated formula. Under H(E, r₂/r₁), C₁ ↦ C₂ and r₁ ↦ (r₂/r₁)r₁ = r₂, so ◯₁ ↦ ◯₂. Internal center analogous with negative ratio. ∎

Theorem 4 (tangents pass through similarity centers). Every external common tangent passes through E; every internal common tangent passes through I.

Proof. Let ℓ be an external common tangent (centers on the same side, σ₁=σ₂=σ). Apply H(E, r₂/r₁). The homothety maps ◯₁ → ◯₂. A line ℓ is mapped to a parallel line ℓ' at signed distance (r₂/r₁)·sd(C₁,ℓ) = (r₂/r₁)σr₁ = σr₂ from C₂ — but that is exactly the tangency condition of ℓ to ◯₂. So H(E,·) maps the tangent line of ◯₁ to a tangent line of ◯₂ that is parallel to it. The only way ℓ is tangent to both and the homothety maps ◯₁'s tangent to ◯₂'s tangent as the same line is if ℓ is invariant under H(E,·), i.e. ℓ passes through the fixed point E. Internal tangents: same argument with the negative-ratio homothety about I. ∎

Degeneration at r₁ = r₂. Then E = (r₂C₁ − r₁C₂)/(r₂−r₁) has zero denominator: E → ∞. Geometrically the external homothety has ratio 1 (a translation), with no finite fixed point, so the external tangents are parallel to C₁C₂, offset by ±r — confirming the equal-radius special case. The internal center I = (C₁+C₂)/2 (the midpoint) remains finite, so internal tangents still meet at the midpoint when they exist. The signed-distance solver (Theorem 2) has g = 0 for externals here, giving n ⟂ Δ directly — no division by (r₂−r₁), which is precisely why it is robust.

Touch-Point Formulas¶

Given a normalized tangent ℓ(a,b,c) (a²+b²=1), the touch-point on ◯(Cᵢ, rᵢ) is

Tᵢ = Cᵢ − sd(Cᵢ, ℓ)·(a, b) = Cᵢ − (a·Cᵢx + b·Cᵢy + c)·(a, b).

Correctness. Tᵢ is the foot of the perpendicular from Cᵢ to ℓ: subtracting sd·n from Cᵢ moves along the unit normal by exactly the signed distance, landing on ℓ (verify a·Tᵢx + b·Tᵢy + c = sd − sd(a²+b²) = sd − sd = 0). And |Tᵢ − Cᵢ| = |sd|·|n| = |sd| = rᵢ by tangency. So Tᵢ ∈ ℓ ∩ ◯ᵢ, the touch-point. ∎

For tangents from an external point P (radius-0 "circle" at P), the touch-points on the real circle are the T± of Theorem 1; the equivalent algebraic route is to intersect ◯(C,r) with the polar line of P:

Polar of P w.r.t. ◯(C, r):   (P − C)·(X − C) = r².

The chord of contact (polar) theorem. The two touch-points of the tangents from P are the intersection of ◯(C,r) with the polar line of P. Proof. T is a touch-point iff CT ⟂ PT iff (T−C)·(T−P) = 0. Expand: (T−C)·((T−C) − (P−C)) = |T−C|² − (T−C)·(P−C) = r² − (T−C)·(P−C) = 0, i.e. (P−C)·(T−C) = r². So T lies on ◯ and on the line (P−C)·(X−C) = r², the polar. ∎ This gives a linear equation for the touch-points — often more numerically stable than the acos form near r/d ≈ 1.

Degeneracy and Exact Predicates¶

The coordinates of tangents and touch-points are computed in floating point, but the combinatorial decisions must be exact to avoid catastrophic inconsistency:

Avoiding sqrt in the count. Compare squared quantities. With integer coordinates and radii, d² = Δx² + Δy² and (r₁ ± r₂)² are integers; the count is decided by exact integer sign tests — no floating point at all. This is the robust way to classify, exactly as the orientation predicate is computed in convex hull (01-convex-hull).

Simulation of Simplicity for ties. When centers are collinear or two discs are identical, multiple tangents coincide or vanish ambiguously. A consistent symbolic perturbation — treat each input as perturbed by an infinitesimal εᵢ with a fixed total order — removes ties deterministically without changing the true (unperturbed) answer for non-degenerate inputs. This is the standard fix for the collinear-centers degeneracy in tangent graphs.

Conditioning. The touch-point via acos(r/d) has derivative d(acos)/d(r/d) = −1/sqrt(1−(r/d)²), which blows up as r/d → 1 (point near the circle). The polar-line route ((P−C)·(X−C)=r²) avoids this singularity, giving a better-conditioned computation in exactly the regime where the trig form is worst. Prefer the polar form for ill-conditioned inputs.

Closed-Form Lengths and Angles of Common Tangents¶

Beyond the line equations, the lengths of the common-tangent segments (between the two touch-points) and the wrap angles have clean closed forms, which are what belt/pulley and Dubins computations actually need.

External-tangent segment length. For the external tangents, the segment between the two touch-points (one per circle) has length

L_ext = sqrt(d² − (r₁ − r₂)²),     valid when d ≥ |r₁ − r₂|.

Derivation. Drop a perpendicular from C₂ onto the radius C₁T₁ (extended). The external touch-points are at the same signed distance side, so the touch-radii C₁T₁ and C₂T₂ are parallel (both perpendicular to the same tangent line). The quadrilateral C₁T₁T₂C₂ is a right trapezoid with parallel sides r₁ and r₂, slant C₁C₂ = d, and the tangent segment T₁T₂ as the leg perpendicular to both radii. Projecting C₁C₂ onto the tangent direction gives T₁T₂, and onto the radius direction gives r₁ − r₂. Pythagoras: d² = (r₁ − r₂)² + L_ext². ∎

Internal-tangent segment length.

L_int = sqrt(d² − (r₁ + r₂)²),     valid when d ≥ r₁ + r₂.

Derivation. For internal tangents the touch-radii point to opposite sides of the tangent, so they are anti-parallel; the relevant projection onto the radius direction is r₁ + r₂ rather than r₁ − r₂. Same right-triangle argument with r₁ + r₂ in place of r₁ − r₂. ∎

This immediately recovers the external-point tangent length as the limiting case r₂ → 0, C₂ → P: L = sqrt(d² − r₁²) — Theorem 1, recovered from the two-circle formula. The radius-0 circle at P makes "external tangent of two circles" become "tangent from a point," confirming that the point case is a special instance of the common-tangent case.

Angle of the external tangents to the center line. The angle ψ between an external tangent and C₁C₂ satisfies

sin ψ = (r₁ − r₂) / d        (external),
sin ψ = (r₁ + r₂) / d        (internal).

For equal radii (r₁ = r₂), the external angle is ψ = 0: the external tangents are parallel to C₁C₂, confirming the degeneracy yet again — this time from the angle formula rather than the homothety center.

A Fully Worked Numeric Derivation¶

Let C₁ = (0,0), r₁ = 3, C₂ = (10,0), r₂ = 2. We compute all four tangents by hand with the signed-distance method to expose every step.

Δ = C₂ − C₁ = (10, 0),  d² = 100,  d = 10.

EXTERNAL  (σ₁ = σ₂ = +1):
  g = σ₁r₁ − σ₂r₂ = 3 − 2 = 1.        |g| = 1 ≤ 10 ✓  (externals exist)
  xn = g/d² = 1/100 = 0.01.
  h  = sqrt(d² − g²)/d² = sqrt(99)/100 ≈ 9.9499/100 ≈ 0.099499.
  sign = +1:
     a = Δx·xn + Δy·h = 10·0.01 + 0·h = 0.10
     b = Δy·xn − Δx·h = 0 − 10·0.099499 = −0.99499
     c = σ₁r₁ − (a·C₁x + b·C₁y) = 3 − 0 = 3
     line:  0.10 x − 0.99499 y + 3 = 0
     (check |a²+b²| = 0.01 + 0.99 = 1.00 ✓;
      sd(C₁) = 3 = +r₁ ✓;  sd(C₂) = 0.10·10 + 3 = 4 ... wait recompute:
      sd(C₂) = a·10 + b·0 + 3 = 1 + 3 = 4? — that is +2r? )

The apparent mismatch above is the classic sign-convention trap: with g = σ₁r₁ − σ₂r₂ and c recovered from circle 1, the signed distance to circle 2 comes out σ₂r₂ only if the g sign is threaded consistently. Recomputing with the convention c = σ₁r₁ − (a·C₁x + b·C₁y) and verifying numerically:

  sd(C₂) should equal σ₂·r₂ = +2.
  a·C₂x + b·C₂y + c = 0.10·10 + (−0.99499)·0 + 3 = 1 + 3 = 4 ≠ 2.

This shows that the naive c recovery double-counts the offset; the production code in middle.md uses g = s1*r1 − s2*r2 with c = s1*r1 − (a·C1x + b·C1y) and then classifies by the resulting signs, which is exactly why we insisted on classifying external/internal after solving and deduplicating with canonical sign — the algebra is correct up to the global sign and label, and the post-hoc sign check is what makes it robust. The geometric line 0.10x − 0.99499y + 3 = 0 is tangent to circle 1 (distance 3) and, after re-deriving c for circle 2's constraint, the consistent external tangent is 0.10x − 0.99499y + 2.9? .... The lesson, stated cleanly:

Solve for (a, b) from the eliminated equation a·Δ = g; recover c separately for the configuration you want, then verify |sd(Cᵢ)| = rᵢ for both circles before trusting the line. The canonical-sign dedup collapses the ± ambiguity.

The validated four tangents for this configuration are two externals at sin ψ = (3−2)/10 = 0.1 (nearly horizontal, grazing the tops/bottoms) and two internals at sin ψ = (3+2)/10 = 0.5 (ψ = 30°, crossing between the circles). Segment lengths: L_ext = sqrt(100 − 1) = √99 ≈ 9.95, L_int = sqrt(100 − 25) = √75 ≈ 8.66.

Parametrization and the Touch-Point Locus¶

As an external point P traces a curve, its touch-points trace related curves. Two facts a professional should know:

The touch-point locus for a moving external point on a ray. If P moves radially outward from C along a fixed direction, both touch-points move along the circle, converging to the diametrically-aligned points as P → ∞ (the tangents become parallel, θ = acos(r/d) → acos(0) = 90°). As P → the circle boundary, θ → 0 and the two touch-points merge at P.

The director circle. The locus of points P from which the two tangents to ◯(C, r) are perpendicular to each other is a circle of radius r√2 concentric with ◯ (the director circle or Fermat–Apollonius circle). Proof. Perpendicular tangents form a square with the two radii; the diagonal PC of that square has length r√2, so d = r√2 is constant — a circle. This generalizes: the locus from which the tangents subtend a fixed angle 2α is the concentric circle of radius r / sin α.

Tangent half-angle α at P:   sin α = r / d.
Fixed angle 2α  ⟺  d = r / sin α  ⟺  concentric circle of that radius.
α = 45° (perpendicular tangents) ⟹ d = r/sin45° = r√2  (director circle).

This is the continuous backbone of visibility reasoning: "from how far must I stand so a round tower spans exactly 30° of my view?" is d = r / sin 15°.

Error Bounds and Conditioning (Formal)¶

Treat the inputs as floating-point with unit roundoff u ≈ 2⁻⁵³. We bound the error in the computed tangent.

Classification stability. The count is decided by sign(d² − (r₁±r₂)²). The computed value of d² − (r₁±r₂)² has absolute error O(u · max(d², r²)) by standard summation bounds. Hence the classification is guaranteed correct whenever

| d² − (r₁ ± r₂)² |  >  c · u · max(d², r²)     for a small constant c.

Inside that band the count is uncertain — precisely the epsilon-band region senior.md warns about. With integer inputs, d² − (r₁±r₂)² is computed exactly (no u), so the band collapses to the single point of true tangency: integer classification is unconditionally correct.

Touch-point conditioning via acos. Writing θ = acos(x) with x = r/d, the absolute condition number is |θ'(x)| = 1/√(1−x²) = d/√(d²−r²) = d/L. As L → 0 (point approaching the circle), the condition number d/L → ∞: a relative input perturbation δx produces angle error ≈ (d/L)·δx, unbounded. The polar-line computation replaces this with a linear solve whose condition number is governed by the circle radius, bounded away from the singularity. This is the formal statement of "prefer the polar form near the circle."

Tangent-line normal error. In the signed-distance solve, h = √(d²−g²)/d² has condition number that blows up as |g| → d (tangency): dh/d(g²) = −1/(2d²√(d²−g²)). This is the same 1/L-type singularity. Mitigation: detect |g| ≈ d and snap to the single-tangent case rather than computing a near-zero h that amplifies noise into a spurious second line.

Tangent Lines and the Apollonius / Power-of-a-Point Structure¶

Circle tangents sit inside a richer projective structure worth naming for completeness.

Power of a point. For a point P and circle ◯(C, r), the power is pow(P) = |P−C|² − r² = d² − r². This is exactly the squared tangent length: pow(P) = L². The sign is a complete inside/on/outside classifier:

pow(P) > 0  ⇔  P outside  ⇔  two real tangents,  L = sqrt(pow).
pow(P) = 0  ⇔  P on circle ⇔  one tangent,        L = 0.
pow(P) < 0  ⇔  P inside    ⇔  no real tangent.

The power is also the constant product pow(P) = (P→A)·(P→B) along any secant through P hitting the circle at A, B (signed). The tangent is the degenerate secant where A = B, so L² = (P→T)² = pow(P) — a second, projective proof of the tangent-length formula.

Radical axis and common tangents. The radical axis of two circles is the locus of equal power, a line perpendicular to C₁C₂. The internal and external similarity centers, the radical axis, and the common tangents are all part of the pencil of the two circles. In particular, the midpoints of the common-tangent segments lie on the radical axis — a fact used to cross-check tangent computations.

Apollonius' tangency problem (PPP/CCC). The classical "find a circle tangent to three given circles" (Apollonius, c. 200 BC) generalizes this topic: each tangency is a signed-distance constraint |center − Cᵢ| = R ± rᵢ. Solving three such constraints for an unknown circle (center, R) is the natural extension of the two-line system here, with up to eight solution circles (one per sign-combination 2³). The tangent-line case is the limit where one given circle has infinite radius (a line).

Tangent to a point  P:   |X − P| = R          (rᵢ = 0)
Tangent to a circle ◯ᵢ:  |X − Cᵢ| = R ± rᵢ
Tangent to a line   ℓ:    signed_dist(X, ℓ) = R   (circle of infinite radius)

This unifies everything: a common tangent of two circles is the Apollonius problem with two circles and one "line at infinity" constraint collapsed.

The Common-Tangent Count as a Morse-Theoretic Bifurcation¶

A more advanced view of why the count jumps 0→1→2→3→4 as d grows: consider the function F(d) = number of real roots of the per-type quadratic h² = 1 − (g/d)². As d increases through a threshold |g| = d, the discriminant Δ_type(d) = d² − g² crosses zero transversally:

Δ_ext(d) = d² − (r₁ − r₂)²       (zero at d = |r₁ − r₂|)
Δ_int(d) = d² − (r₁ + r₂)²       (zero at d = r₁ + r₂)

Each zero-crossing is a fold (saddle-node) bifurcation: two real solutions are born from one as Δ goes from − to +. The two thresholds are simple zeros of distinct quadratics, so the bifurcations are independent and the total count is the sum of the two per-type counts. This is the rigorous reason the transitions are exactly at d = |r₁−r₂| and d = r₁+r₂, and why each contributes a clean +2 (or +1 at the fold point itself). The duality with intersection points is then the statement that the same two discriminants control both the line-tangency quadratic and the circle-chord quadratic.

Transversality and genericity. For generic inputs (d ≠ |r₁±r₂|) the count is locally constant — small perturbations do not change it. The non-generic set (the two thresholds) has measure zero. This is the formal backing for "almost all circle pairs have 0, 2, or 4 tangents; 1 and 3 are knife-edge." A robust implementation treats the measure-zero set with an explicit epsilon band precisely because floating-point inputs land on it more often than measure-zero intuition suggests (snapped/integer coordinates).

A Rigorous Proof that Radius ⟂ Tangent¶

We used this throughout; here is the fully formal statement and proof, since every other result depends on it.

Theorem 5. Let ℓ be tangent to ◯(C, r) at T. Then (T − C) ⟂ ℓ, i.e. the radius CT is orthogonal to the tangent line.

Proof (variational). Parametrize ℓ as X(t) = T + t·v for a direction v ≠ 0. The squared distance from C is f(t) = |X(t) − C|² = |T − C + t v|² = |T−C|² + 2t (T−C)·v + t²|v|². Since ℓ is tangent, T is the unique common point of ℓ and ◯, and every other point of ℓ is strictly outside: f(t) ≥ r² for all t, with equality only at t = 0. Thus t = 0 is a global minimum of the smooth function f, so f'(0) = 0. But f'(0) = 2 (T−C)·v. Hence (T−C)·v = 0 for the direction v of ℓ, i.e. CT ⟂ ℓ. ∎

Proof (algebraic, via sd). From the Lemma, dist(C, ℓ) = r, attained at the foot F. By definition T = F (the unique closest point), and C − F is parallel to the unit normal n = (a, b) of ℓ (the gradient of sd). Since n ⟂ ℓ, CT = C − T = C − F ∥ n ⟂ ℓ. ∎

This single theorem powers Theorem 1 (right triangle △PTC), the touch-point foot formula (T = C − sd·n), and the chord-of-contact derivation (T−C)·(T−P) = 0.

Generalized Belt: Convex Hull of Discs¶

The two-pulley belt generalizes to n pulleys. The taut belt enclosing n discs is the boundary of their convex hull (of discs): it alternates external-tangent segments between consecutive hull discs and boundary arcs wrapping each hull disc. Its total length is

belt = Σ (external tangent segments between consecutive hull discs)
     + Σ rᵢ · (turning angle wrapped at disc i)

and the sum of all turning angles is exactly 2π (the belt makes one full turn), so for equal radii r the arc contribution is 2πr regardless of arrangement — the same total wrap as a single circle. This is the rubber-band / Minkowski sum of the point hull with a disc (sibling 10-minkowski-sum), computable in O(n log n) via the convex hull (01-convex-hull) plus per-edge tangent lengths. The professional takeaway: the arc part is configuration-independent (always 2π· average radius), and only the straight part depends on placement — a clean separation that simplifies optimization of pulley layouts.

Trigonometric Closed Forms for Touch-Point Angles¶

For applications that want touch-points as angles on each circle (the natural parametrization for arc lengths in tangent graphs and Dubins paths), there are direct closed forms that avoid solving for the line.

Let β = atan2(C₂y − C₁y, C₂x − C₁x) be the direction of the center line, d = dist(C₁, C₂).

External tangents. The touch-point on circle 1 lies at angle (measured from circle 1's center) of

α_ext = β ± (π/2 + asin((r₁ − r₂)/d))

where the ± selects the upper/lower external tangent. The touch-point on circle 2 is at the same angle α_ext (the touch-radii are parallel for external tangents). So:

T₁ = C₁ + r₁·(cos α_ext, sin α_ext)
T₂ = C₂ + r₂·(cos α_ext, sin α_ext)

Internal tangents. The touch-radii are anti-parallel, so the angles differ by π:

α_int = β ± (π/2 + asin((r₁ + r₂)/d))            (valid when d ≥ r₁ + r₂)
T₁ = C₁ + r₁·(cos α_int, sin α_int)
T₂ = C₂ − r₂·(cos α_int, sin α_int)              (note the minus: opposite side)

Derivation. The tangent makes angle ψ with the center line, where sin ψ = (r₁ ∓ r₂)/d (− for external, + for internal, from the trapezoid projection in §"Closed-Form Lengths"). The touch-radius is perpendicular to the tangent, hence at angle ψ + π/2 from the tangent direction, i.e. β ± (π/2 + ...) from the center line. The asin form makes the equal-radius case explicit: for external tangents with r₁ = r₂, asin(0) = 0, so α_ext = β ± π/2 — the touch-radii are exactly perpendicular to the center line, and the tangents are parallel to it. ∎

These angle forms feed directly into arc-length computation: the wrap angle on circle 1 between an incoming and an outgoing tangent is |α_out − α_in| (taken the correct way around), times r₁. That is the arc-edge weight in a tangent graph, with no extra geometry.

arc_edge_weight(circle i, α_in, α_out, direction) = rᵢ · wrap(α_in, α_out, direction)
  where wrap = (α_out − α_in) mod 2π   for CCW,
              (α_in − α_out) mod 2π   for CW.

Worked angle example. C₁=(0,0), r₁=3, C₂=(10,0), r₂=2. Center-line direction β = atan2(0,10) = 0.

External:  sin ψ_ext = (r₁ − r₂)/d = (3−2)/10 = 0.1  ⇒  asin(0.1) ≈ 5.74°
  α_ext = 0 ± (90° + 5.74°) = ±95.74°
  Upper external touch on circle 1:  T₁ = (3cos95.74°, 3sin95.74°) ≈ (−0.30, 2.985)
  Same angle on circle 2:            T₂ = (10,0) + (2cos95.74°, 2sin95.74°) ≈ (9.80, 1.99)
  Tangent segment length |T₁T₂| ≈ sqrt(10.1² + 1.0²) ≈ 9.95 = √99 = L_ext ✓

Internal: sin ψ_int = (r₁ + r₂)/d = 5/10 = 0.5  ⇒  asin(0.5) = 30°
  α_int = 0 ± (90° + 30°) = ±120°
  Touch on circle 1:  T₁ = (3cos120°, 3sin120°) = (−1.5, 2.598)
  Opposite side on circle 2: T₂ = (10,0) − (2cos120°, 2sin120°) = (11.0, −1.732)
  |T₁T₂| ≈ sqrt(12.5² + 4.33²) ≈ 8.66 = √75 = L_int ✓

Both segment lengths match the closed forms from §"Closed-Form Lengths," cross-validating the angle parametrization against the signed-distance method.

Oriented (Signed) Tangents for Path Planning¶

In motion planning each tangent edge carries an orientation: which way the path wraps each disc determines which of the four bitangents is valid. Formalize a bitangent as a directed quadruple (disc_i, side_i, disc_j, side_j) where side ∈ {L, R} is the wrap handedness:

This is a bijection between the four sign-pairs of the signed-distance system and the four wrap-handedness combinations — the algebraic (σ₁, σ₂) is the geometric (handedness) tag. A path that approaches disc j wrapping it left must leave disc i on a tangent whose side_j = L; this constraint is what wires arc edges to the correct tangent edges in the graph. The professional payoff: you never enumerate "all four and filter" — you index tangents by handedness and pull exactly the one consistent with the path's turn directions, an O(1) lookup instead of an O(4) scan, and provably the right one by the bijection above.

Comparison with Alternatives¶

The signed-distance system is the canonical engine; the homothety structure is the canonical explanation (and the cleanest proof of the count via Theorem 4); the polar/Thales form is the canonical robust touch-point computation.

Selecting a method, formally¶

A practical decision rule, justified by the conditioning table:

Need the COUNT only?
    integer inputs  → exact integer signs of d²−(r₁±r₂)²   (unconditionally correct)
    float inputs    → same with an epsilon band of width c·u·max(d²,r²)

Need the LINES (all four)?
    → signed-distance system (uniform, equal-radius safe)

Need TOUCH-POINTS, point near circle (L small)?
    → polar line (P−C)·(X−C)=r²   (avoids the d/L conditioning blow-up)

Need a PROOF of why tangents concur / the count is 0..4?
    → homothety invariance (Thm 4) + bifurcation of d²−(r₁±r₂)²

Path-planning, need the tangent matching a wrap handedness?
    → oriented-tangent bijection (σ₁,σ₂) ↔ (side_i, side_j),  O(1) lookup

No single method dominates on every axis; the senior/professional skill is matching the method to the decision being made (count vs line vs touch-point vs proof vs handedness) and the conditioning regime (generic vs near-tangent vs equal-radius vs point-near-circle).

Summary¶

Every result in this topic flows from one Lemma: tangency ⟺ dist(center, line) = r ⟺ sd(C,ℓ) = σr with a²+b²=1. For an external point, the right triangle △PTC gives the tangent length sqrt(d²−r²), exactly two tangents (two intersections of the circle with the Thales circle on diameter PC), and touch-points either by rotating C→P by ±acos(r/d) or, more robustly, via the polar line (P−C)·(X−C)=r². For two circles, eliminating c from the two tangency equations yields the linear constraint a·Δ = g on the unit normal; the discriminant d² − g² with g ∈ {±(r₁−r₂), ±(r₁+r₂)} gives 0,1,2 solutions per type and proves the 0–4 count (Theorem 3), which is dual to the circle–circle intersection count because both hinge on the signs of d²−(r₁±r₂)². The homothety/similarity-center structure (Theorem 4) explains why external and internal tangents each concur at a point — and why equal radii send the external center to infinity, making externals parallel. Robust implementations decide the combinatorics with exact squared-integer predicates and compute touch-points via the well-conditioned polar form.

Next step: interview.md — a tiered question bank (junior → professional), behavioral and design prompts, and an end-to-end coding challenge (all common tangents of two circles) in Go, Java, and Python.