Mutually Tangent Circles

Problem: The Small Middle Circle

Finding Point A

Let point A be (x, y). Thus:

• A (x, y)

• B (5, 0)

• C (0, 0)

Knowing A is 4 units away from origin:
x² + y² = 4²
x² + y² = 16

Knowing distance AB = 3
(x - 5)²+ (y - 0)² = 3²
x² - 10x + 25 + y² = 9
x² + y² = 10x - 16

Thus,
16 = 10x - 16
x = 3.2

From x² + y² = 16:
3.2² + y² = 16
y = 2.4

Rewriting:

• Point A (3.2, 2.4)

• Point B (5, 0)

• Point C (0, 0)

Finding Point D

Let radius of circle D be r. Let point D be (x, y).

Because circle D touches circle A, whose radius is 1, the distance between their centers is r + 1. Then:
(x - 3.2)² + (y - 2.4)² = (r + 1)²

Similarly, circle D touches circle B, whose radius is 2, the distance between their centers is r + 2. Then:
(x - 5)² + (y - 0)² = (r + 2)²
(x - 5)² + y² = (r + 2)²

Also, circle D touches circle C, whose radius is 3, the distance between their centers is r + 3. Then:
(x - 0)² + (y - 0)² = (r + 3)²
x² + y² = (r + 3)²

We now have three equations and three unknowns - x, y and r.

Expand LHS of (x - 5)² + y² = (r + 2)²:
x² - 10x + 25 + y² = (r + 2)²
x² + y² = (r + 2)² + 10x - 25

Substituting in x² + y² = (r + 3)²:
(r + 3)² = (r + 2)² + 10x - 25
r² + 6r + 9 = r² + 4r + 4 + 10x - 25
2r = 10x - 30
r = 5x - 15
x = (r + 15) / 5
x = r/5 + 3

Expand LHS of (x - 3.2)² + (y - 2.4)² = (r + 1)²:
x² - 6.4x + 10.24 + y² - 4.8y + 5.76 = (r + 1)²
x² + y² = (r + 1)² + 6.4x + 4.8y - 16

Substituting in x² + y² = (r + 3)²:
(r + 3)² = (r + 1)² + 6.4x + 4.8y - 16
r² + 6r + 9 = r² + 2r + 1 + 6.4x + 4.8y - 16
4r = 6.4x + 4.8y - 24
r = 1.6x + 1.2y - 6

Substituting in x = (r + 15) / 5:
r = 1.6 * (r + 15) / 5 + 1.2y - 6
r = 0.32r + 4.8 + 1.2y - 6
0.68r = 1.2y - 1.2
34r / 60 = y - 1
y = 17r/30 + 1

Thus, point D (r/5 + 3, 17r/30 + 1).

Finding Radius r of Circle D

Knowing distance between D and origin is r + 3:
(r/5 + 3)² + (17r/30 + 1)² = (r + 3)²
r²/25 + 6r/5 + 9 + 289r²/900 + 17r/15 + 1 = r² + 6r + 9
23r²/36  + 11r/3 - 1 = 0
23r² + 132r - 36 = 0

Solve quadratic:
r = (-132 ±√( 132² -4(23)(-36) ) ) / (2 * 23)
r = (-132 ±√20736) / 46
r = (-132 ± 144) / 46

Since r > 0:
r = (-132 + 144) / 46
r = 6 / 23

Thus, in between 3 circles of radius 1, 2, 3 units respectively, that tangent to each other, the circle that fits in the middle has a radius of 6 / 23 units.

What about a general solution?

Finding Point A

Let point A be (x, y). Thus:

• A (x, y)

• B (r₂ + r₃, 0)

• C (0, 0)

Knowing A is (r₁ + r₃) away from the origin:
x² + y² = (r₁ + r₃)²

Distance AB = (r₁ + r₂):
(x - (r₂ + r₃))² + (y - 0)² = (r₁ + r₂)²
(x - (r₂ + r₃))² + y² = (r₁ + r₂)²

Subtract the 2 equations from one another:
(x - (r₂ + r₃))² - x² = (r₁ + r₂)² - (r₁ + r₃)²
x² - 2x (r₂ + r₃) + (r₂ + r₃)² - x² = (r₁ + r₂)² - (r₁ + r₃)²
- 2x (r₂ + r₃) + (r₂ + r₃)² = (r₁ + r₂)² - (r₁ + r₃)²
- 2x (r₂ + r₃) + (r₂² + 2r₂r₃ + r₃²) = (r₁² + 2r₁r₂ + r₂²) - (r₁² + 2r₁r₃ + r₃²)
- 2x (r₂ + r₃) + 2r₂r₃ + r₃² = 2r₁r₂ - 2r₁r₃ - r₃²
- 2x (r₂ + r₃) = 2r₁r₂ - 2r₁r₃ - 2r₂r₃ - 2r₃²
x (r₂ + r₃) = r₃² + r₁r₃ + r₂r₃ - r₁r₂
x (r₂ + r₃) = r₃(r₃ + r₁) + r₂(r₃ - r₁)
x = (r₃(r₁ + r₃) + r₂(r₃ - r₁)) / (r₂ + r₃) 

From x² + y² = (r₁ + r₃)²:
y² = (r₁ + r₃)² - x² 

Let's not expand them to keep manageable. So, Point A = (x, y) where:

• x = (r₃(r₁ + r₃) + r₂(r₃ - r₁)) / (r₂ + r₃) 

• y² = (r₁ + r₃)² - x²

Finding Point D

Let centre of circle D = (u, v) and it has a radius r.

Because circle D touches circle A, whose radius is r₁, the distance between their centers is r + r₁. Then:
(u - x)² + (v - y)² = (r + r₁)² ~ Equation 1

Similarly, circle D touches circle B, whose radius is r₂, the distance between their centers is r + r₂. Then:
(u - (r₂ + r₃))² + (v - 0)² = (r + r₂)²
(u - (r₂ + r₃))² + v² = (r + r₂)² ~ Equation 2

Also, circle D touches circle C, whose radius is r₃, the distance between their centers is r + r₃. Then:
(u - 0)² + (v - 0)² = (r + r₃)²
u² + v² = (r + r₃)² ~ Equation 3

Equation 2 - Equation 3:
(u - (r₂ + r₃))²  - u² = (r + r₂)² - (r + r₃)²
u² - 2u(r₂ + r₃) + (r₂ + r₃)² - u² = r² + 2rr₂ + r₂² - r² - 2rr₃ - r₃²
- 2u(r₂ + r₃) + (r₂ + r₃)² = 2rr₂ + r₂² - 2rr₃ - r₃²
- 2u(r₂ + r₃) + (r₂ + r₃)² = 2r(r₂ - r₃) + r₂²- r₃²
(r₂ + r₃)² - (r₂²- r₃²) = 2r(r₂ - r₃) + 2u(r₂ + r₃)

Since (r₂ + r₃)² - (r₂²- r₃²) = 2r₂r₃ + 2r₃²:
(r₂ + r₃)² - (r₂²- r₃²)  = 2r₃(r₂ + r₃)

Substituting back:
2r₃(r₂ + r₃) = 2r(r₂ - r₃) + 2u(r₂ + r₃)
r₃(r₂ + r₃) = r(r₂ - r₃) + u(r₂ + r₃)
u = (r₃(r₂ + r₃) + r(r₃ - r₂)) / (r₂ + r₃)

Equation 1 - Equation 3:
(u - x)² + (v - y)² - (u² + v²) = (r + r₁)² - (r + r₃)²
u² - 2ux + x² + v² - 2vy + y² - u² - v² = r² + 2rr₁ + r₁² - r² - 2rr₃ - r₃²
- 2ux + x² - 2vy + y² = 2rr₁ + r₁² - 2rr₃ - r₃²
x² + y² - 2ux - 2vy = 2r(r₁ - r₃) +  r₁² - r₃²

Substituting in x² + y² = (r₁ + r₃)²:
(r₁ + r₃)² - 2ux - 2vy = 2r(r₁ - r₃) +  r₁² - r₃²
r₁² + 2r₁r₃ + r₃² - 2ux - 2vy = 2r(r₁ - r₃) +  r₁² - r₃²
2r₁r₃ + 2r₃² - 2ux - 2vy = 2r(r₁ - r₃)
r₃(r₁ + r₃) - ux - vy = r(r₁ - r₃)
ux + vy = r₃(r₁ + r₃) + r(r₃ - r₁)
v = (r₃(r₁ + r₃) + r(r₃ - r₁) - ux) / y

Thus, D = (u, v), where:

• u = (r₃(r₂ + r₃) + r(r₃ - r₂)) / (r₂ + r₃)

• v = (r₃(r₁ + r₃) + r(r₃ - r₁) - ux) / y

Omelette 'em All

This is where it tends to get ugly fast.

1. Eliminate v

The goal is to clear all unknowns except r, r₁, r₂, r₃ remain. With these known:

• u² + v² = (r + r₃)²

• x = (r₃(r₁ + r₃) + r₂(r₃ - r₁)) / (r₂ + r₃)

• y² = (r₁ + r₃)² - x²

• u = (r₃(r₂ + r₃) + r(r₃ - r₂)) / (r₂ + r₃)

• v = (r₃(r₁ + r₃) + r(r₃ - r₁) - ux) / y

Rearrange v = (r₃(r₁ + r₃) + r(r₃ - r₁) - ux) / y:
vy = (r₃(r₁ + r₃) + r(r₃ - r₁) - ux)

Square both sides:
v²y² = (r₃(r₁ + r₃) + r(r₃ - r₁) - ux)²

Substitute in y² = (r₁ + r₃)² - x²:
v²((r₁ + r₃)² - x²) = (r₃(r₁ + r₃) + r(r₃ - r₁) - ux)²

Substitute in v² = (r + r₃)² - u²:
((r + r₃)² - u²) * ((r₁ + r₃)² - x²) = [r₃(r₁ + r₃) + r(r₃ - r₁) - ux]²

This is the key equation.

2. Make Substitution Less Ugly

Let:

• D = (r₂ + r₃)

• X = r₃(r₁ + r₃) + r₂(r₃ - r₁)

• U = r₃(r₂ + r₃) + r(r₃ - r₂)

So, x = X / D, u = U / D.

3. Substitute x = X / D into ((r₁ + r₃)² - x²) of the Key Equation

((r₁ + r₃)² - x²) = ((r₁ + r₃)² - (X / D)²)
((r₁ + r₃)² - x²) = ((r₁ + r₃)² - X² / D²)
((r₁ + r₃)² - x²) = ((D²(r₁ + r₃)² - X²) / D²)

4. Substitute u = U / D into ((r + r₃)² - u²) of the Key Equation

((r + r₃)² - u²) = ((r + r₃)² - (U / D)²)
((r + r₃)² - u²) = ((r + r₃)² - U² / D²)
((r + r₃)² - u²) = ((D²(r + r₃)² - U²) / D²)

5. Substitute u and x into the RHS of the Key Equation

The RHS of the key equation being: [r₃(r₁ + r₃) + r(r₃ - r₁) - ux]²
[r₃(r₁ + r₃) + r(r₃ - r₁) - ux]² = [r₃(r₁ + r₃) + r(r₃ - r₁) - UX / D²]²

6. Put everything into the Key Equation

We have:

• ((r₁ + r₃)² - x²) = ((D²(r₁ + r₃)² - X²) / D²)

• ((r + r₃)² - u²) = ((D²(r + r₃)² - U²) / D²)

• [r₃(r₁ + r₃) + r(r₃ - r₁) - ux]² = [r₃(r₁ + r₃) + r(r₃ - r₁) - UX / D²]²

With ((r + r₃)² - u²) * ((r₁ + r₃)² - x²) = [r₃(r₁ + r₃) + r(r₃ - r₁) - ux]²:
((D²(r + r₃)² - U²) / D²) * ((D²(r₁ + r₃)² - X²) / D²) = [r₃(r₁ + r₃) + r(r₃ - r₁) - UX / D²]²
( (D²(r + r₃)² - U²) * (D²(r₁ + r₃)² - X²) ) / D⁴ = [r₃(r₁ + r₃) + r(r₃ - r₁) - UX / D²]²
(D²(r + r₃)² - U²) * (D²(r₁ + r₃)² - X²) = D⁴ [r₃(r₁ + r₃) + r(r₃ - r₁) - UX / D²]²
(D²(r + r₃)² - U²) * (D²(r₁ + r₃)² - X²) = [D²(r₃(r₁ + r₃) + r(r₃ - r₁)) - UX]²

The Big Circle E