Mutually Tangent Circles

This article will examine circles that are mutually tangent to each other with distinct points of tangency. We build up from two, to three, then to four mutually tangent circles. This will prepare the way for the next article on Descartes' Circle Theorem.

Let R denote the set of real numbers, R⁺ the non-negative real numbers, RxR = R² = {X=(x,y): x∈ R,y ∈R} the real plane. That is R² is the set of all ordered pairs of real numbers.

A relation in R² is any subset of R². Thus a line, {X:y=ax+b}, and a circle, {X: XA=r}, are relations in R², the latter having radius r and center A.

Two circles are tangent to each other (mutually tangent) if the relations defining them have exactly one point in common.

Two Tangent Circles

Given circle E_A = {X:XA = r_A}, find circle E_C = {X:XC = r_C} tangent to E_A. There are two cases. Case A: E_A surrounds E_C ; Case B: E_A and E_C are externally tangent to each other.

• Case A: Let C ∈ {X: XA < r_A}, i.e. C is in the interior of E_A. Then E_C = {X: XC = r_A-AC} is tangent to E_A. Proof: First we show they have a point in common. Let the radial segment from A through C hit E_A at P. Then P ∈ E_A and r_A =AP=AC+CP. So CP=r_A-AC and P ∈ E_C. Now we show P is unique. Suppose there is another point Q ≠ P such that Q ∈ E_A and E_C. Then AQP form a triangle. AQ = r_A and CQ = r_A - AC. Thus CQ = AQ-AC. But in triangle ΔACQ we must have CQ + AC > AQ. Thus there is no such Q. That is P is the unique intersection of the two circles. Note: In particular, the line joining the centers A and C includes the point of tangency, P.
• Case B: Let C ∈ {X: XA > r_A}, i.e. C is in the exterior of E_A. Then E_C = {X: XC = AC-r_A} is tangent to E_A. Proof: Similar to case A. Draw the segment AC and let it intersect E_A at P. Then PC=AC-r_A, so P is also in E_A. Again suppose there is another point Q ≠ P such that Q ∈ E_A and E_C. If Q is collinear with A,P, and C then QC = r_A + AC so Q ∉ E_C. In ΔAQC, AQ + QC > AC so QC > AC - AQ = AC - r_A, so Q ∉ E_C. Thus P is the unique point of intersection of the two circles so they are mutually tangent.

Three Mutually Tangent Circles with Three Distinct Points of Tangency

For each of the cases of two tangent circles we now look at how to include a third circle which is mutually tangent to the two at three distinct points of tangency. In Case A, in order for the points of tangency to be distinct, the new circle E_B must be surrounded by E_A. In Case B, E_B might surround E_A and E_C or it might be externally tangent, to both of them.

• Case A: We can find the third tangent circle in two ways. Case A1: Choose any point B ∈ {X:XA < r_A} ∩ {X:XC > r_C} as the center of E_B and then find r_B. Case A2: Choose any 0 < r_B < r_A-r_C, the range of possible radius measurements for circles with centers inside E_A and outside E_C, and then find the center B.
• Case B: We have two cases and each can be done in two ways (as with Case A). Case B1: E_B surrounds E_A and E_C. Case B2: E_A, E_B, and E_C are externally tangent to one another. We divide these into four cases. Case B11: Choose B in the exterior of both E_A and E_C then find r_B. Case B12: Choose a radius 0 < r_B and find a center B. Case B21: Choose B in the exterior of both E_A and E_C then find r_B. Case B22: Choose a radius 0 < r_B and find a center B.

First we examine cases A1, B11, and B21 where we are given A, r_A, C, r_C and B and we need to find r_B. Note, we will also get expressions for r_A and r_C which will be useful throughout. In each case we consider ΔABC with sides of length a=BC, b=AC, and c=AB and semi-perimeter s = (a + b + c)/2.

• Case B11: We can write r_B = c + r_A = a + r_C and b = r_A + r_C. Thus a+ B + c = 2s = 2r_B. Thus
  • r_B = s
  • r_A = s-c
  • r_C = s-a.
• Case B21: Using the fact that the segment joining the centers of two tangent circles includes the point of tangency (from the analysis of two tangent circles), we have a = r_B + r_C, b = r_A + r_C, and c = r_A + r_B. Thus (a + b + c) = 2s = 2(r_A + r_B + r_C). Therefore,
  • r_B = s - (r_A + r_C) = s-b
  • r_A = s-a
  • r_C = s-c.
• Here is a picture of the two constructions using A(0,0), C(3,0) and B(1,1). The black trio is case B11 and the orange trio is case B21.
• Case A1: Choose B in the interior of E_A and the exterior of E_C. From Case B11 above we know that r_B = s-c.

Next we will look at the cases A2, B21, and B22 where we are given A, r_A, C, r_C and r_B and we need to find B. In each case we will set up a system of two quadratic equations parameterized with a linear function of r_B which we can solve for B.

• Case A2: B must satisfy BA = r_A - r_B and BC = r_B + r_C. Notice that E_B is at its maximum radius when B is collinear with A and C. In this case 2r_B + 2r_C = 2r_A. Thus the maximum value for r_B is r_A - r_C.
• Case B12: E_B surrounds. B must satisfy BA = r_A - r_B and BC = r_B - r_C.
• Case B22: E_A, E_B, E_C externally tangent. B must satisfy BA = r_A + r_B and BC = r_B + r_C.

Since A and C are arbitrary, we assume A is the origin of a rectilinear coordinate system, so A = (0,0). Also without loss of generality we can take C = (b,0) so that AC = b in ΔABC. Then our system, in general, becomes

• x² + y² = ƒ₁²(r_B)
• (x-b)² + y² = ƒ₂²(r_B)

Solve in terms of parameter r_B by subtracting the second equation from the first to get 2bx - b² = ƒ₁² - ƒ₂² or,

x = (b² + ƒ₁² - ƒ₂²)/2b and y = ±√[ƒ₁² - x²]. We get the solutions for our three cases by specifying b, ƒ₁ and ƒ₁for each.

• Case A2: b = (r_A - r_C); ƒ₁(r_B) = r_A - r_B; ƒ₂(r_B) = r_C + r_B, 0 < r_B < r_A - r_C.
Case B12: b = (r_A - r_C); ƒ₁(r_B) = r_A - r_B; ƒ₂(r_B) = r_B - r_C.
• Case B22: b = (r_A - r_C); ƒ₁(r_B) = r_A + r_B; ƒ₂(r_B) = r_C + r_B.

In each case we get two solutions for B, one above the line through A and C and one below. We must have r_A ≠ r_B to get a solution for B.

Four Mutually Tangent Circles with Six Distinct Points of Tangency

From the last section, we have two distinct cases of three mutually tangent circles with distinct points of tangency to extend to a fourth circle. Case A: One of the three surrounds the other two. Case B: the three circles are externally tangent to each other. In these cases, to add a 4th tangent circle, we will get systems of three quadratic equations in three unknowns, the coordinates of the center D and the radius d of the fourth tangent circle E_D. As in the case of three circles, the difference in the systems are the linear functions of d in those systems. We will call A = (0,0), C(b₁,0), and B(c₁,c₂). The sides of ΔABC are then a = √[(c₁-b₁)² + c₂²], b = b₁, and c = √[c₁² + c₂²]. Again let s = (a + b + c)/2.

• Case A:

  In this case E_A surrounds the other three circles and there are two placements for E_D; one on the same side of the line through B and C as A and one on the far side of this line from A.

  • AD = ƒ₁(d) = r_A - d
  • BD = ƒ₂(d) = r_B + d
  • CD = ƒ₃(d) = r_C + d
• Case B:

  In the case that E_D surrounds E_A, E_B, and E_C, then AD = ƒ₁(d) = d - r_A; BD = ƒ₂(d) = d - r_B; CD = ƒ₃(d) = d - r_C, where d is the radius of E_D. In particular, AD = ƒ₁(d) = | d - r_A|; BD = ƒ₂(d) = | d - r_B|; CD = ƒ₃(d) = | d - r_C| > 0, where d is greater than r_A, r_B, and r_C and d is the radius of E_D.

  In the case that E_D is externally tangent to E_A, E_B, and E_C, then AD = ƒ₁(d) = r_A + d; BD = ƒ₂(d) = r_B + d; CD = ƒ₃(d) = r_C + d, where d is the radius of E_D. We can rewrite these equations as AD = ƒ₁(d) = r_A - (-d); BD = ƒ₂(d) = r_B - (-d); CD = ƒ₃(d) = r_C - (-d) and again as AD = ƒ₁(d) = |(-d) - r_A|; BD = ƒ₂(d) = |(-d) - r_B|; CD = ƒ₃(d) = |(-d) - r_C|, where (-d) < 0.

  Thus in general we have:

  • AD = ƒ₁(d) = | d - r_A|
  • BD = ƒ₂(d) = | d - r_B|
  • CD = ƒ₃(d) = | d - r_C|

  where this d ∈ R-{0}. We will see below that in solving for d we get a quadratic in d which yields solutions D₁, D₂ such that if D_i < 0 then | D_i| is the radius of an externally tangent E_D and if D_i > 0 then D_i is the radius for E_D that surrounds the other three circles.

Now consider the general solution (for both cases A and B) involving the general ƒ₁(d), ƒ₂(d), ƒ₃(d), and then look at the differences when the actual values of these functions are inserted. For ease of notation we will use f = ƒ₁, g = ƒ₂, h = ƒ₃(d). The general problem is to solve the system:

• E_A(d): x² + y² = f²(d)
• E_B(d): (x-c₁)² + (y-c₂)² = g²(d)
• E_C(d): (x-b₁)² + y² = h²(d)

We solve the system in three steps.

• Find the line of intersection for E_A(d) and E_C(d), L_AC(d), as a function of d.
• Find the line of intersection for E_A(d) and E_B(d), L_AB(d), as a function of d.
• Find the points where the lines of intersection intersect our circles of interest, say E_A(d).
• The lines of intersection represent the points where the distances to E_A(d) and E_C(d) or E_A(d) and E_B(d) are the same for a given d. We want to find d where all three distances are the same for a given d. That is, we want to find the value(s) of d where the lines L_AC(d) and L_AB(d) intersect. These will give us center points for E_D with radius d.

Find L_AC(d):

• x² + y² - f²(d) = (x-b₁)² + y² - h²(d)
• x = (b₁² + f² - h²) / 2b₁

Find L_AB(d):

• x² + y² - f²(d) = (x-c₁)² + (y-c₂)² - h²(d)
• 2c₁x + 2c₂y = c₁² + c₂² + f² - g² = c² + f² - g².
• Letting u = (c² + f² - g²) / 2c₂, we have y = (-c₁/c₂)x + u.

Find the intersections with E_A(d).

• L_AC(d) meets E_A(d) at x₁(d) = (b₁² + f² - h²) / 2b₁ and y₁(d) = ±√[f² - x₁²(d)].
• L_AB(d) meets E_A(d) when x² + ((-c₁/c₂)x + u)² = f².
• This leads to the quadratic equation (c/c₂)²x² - (2c₁u/c₂)x + u² -f² = 0
• Use the quadratic formula to get two solutions for x:
• x₂(d) = (c₁c₂u + c₂√[(cf)² - (d₁u)²]) / c²
• x₃(d) = (c₁c₂u - c₂√[(cf)² - (d₁u)²]) / c²
• and the corresponding y terms y₂(d) = (-c₁/c₂)x₂ + u and y₃(d) = (-c₁/c₂)x₃ + u

To find point(s) where all three equations of the system are satisfied we have to find a value(s) for d such that x₁(d) = x₂(d) or x₁(d) = x₃(d). That is where

• (b₁² + f² - h²) / 2b₁ = (c₁c₂u ± c₂√[(cf)² - (d₁u)²]) / c²
• We isolate the radical, square both sides, substitute for u, and then collect terms with common differences of squares of functions f, g, and h to get:
• (f² - h²)²[c/b₁]² + (f² - g²)² - (f² - h²)(f² - g²)[2c₁/b₁] + (f² - h²)[2c²(1-c₁/b₁)] + (f² - g²)[2(c² - c₁b₁)] - 4c₂²f² + c²[b₁² + c² - 2c₁b₁] = 0

Notice that f, g, and h only show up as squares so the absolute value does not come into play when we make the substitutions. By using the values from the beginning of this section we can compute:

• Case A: f² - g² = w(2d-c) and f² - h² = v(2d-b) where w = -(a + b) and v = -(a + c). Also, f² = (s-d)².
• Case B: f² - g² = w(2d-c) and f² - h² = v(2d-b) where w = (a - b) and v = (a - c). Also, f² = (d-(s-a))².

We substitute these values into the equation in the last paragraph and collect terms with the same degree of d to get a quadratic in d: αd² + βd + κ = 0, where the coefficients for each case are as follows. Recall, b = b₁.

• Case A: w = -(a + b) and v = -(a + c)
  • α = 4[(cv/b)² - 2c₁vw/b + w² - c₂²]
  • β = 4[-(cv)²/b + c₁vw(b+c)/b - cw² + vc²(1-c₁/b) + w(c² - bc₁) + 2sc₂²]
  • κ = (cv)² - 2c₁cvw + (cw)² - 2bvc²(1-c₁/b) - 2cw(c² - bc₁) - 4s²c₂² + c²[b² + c² - 2c₁b]
• Case B: w = (a - b) and v = (a - c)
  • α = 4[(cv/b)² - 2c₁vw/b + w² - c₂²]
  • β = 4[-(cv)²/b + c₁vw(b+c)/b - cw² + vc²(1-c₁/b) + w(c² - bc₁) + 2(s-a)c₂²]
  • κ = (cv)² - 2c₁cvw + (cw)² - 2bvc²(1-c₁/b) - 2cw(c² - bc₁) - 4(s-a)²c₂² + c²[b² + c² - 2c₁b]

Notice that besides the values for v and w, the only difference in these cases is the term "s" in Case A is replaced by "s-a" in case B in the β and κ terms. We solve for d = (-β ±√[β² - 4ακ])/2α. Let D₁ be the root with positive radical and let D₂ be the root with the negative radical. We interpret the roots as follows.

• Case A: In this case the roots are positive and each gives the radius of one of the two possible 4th tangent circles.
• Case B: If a root is negative, its absolute value is the radius of an externally tangent fourth circle. If a root is positive it is the radius of fourth tangent circle that surrounds the original three.
• In either case, for i = 1, 2 and j = 2, 3: If x₁(D_i) = x_j(D_i) then (x - x_j(D_i))² + (y - y_j(D_i))² = (D_i)² is a fourth tangent circle to the original three.

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Examples of four mutually tangent circles

The final part of this article is to show some examples of four tangent circles for different values of b₁, c₁, and c₂. Given the ΔABC we can have the three mutually tangent circles E_A, E_B, E_C with either radii s, s-c, s-b (E_A surrounds, case A) or s-a, s-b, s-c (externally tangent, Case B) respectively. We will label our examples below with the notation: [Case,0,0,c₁,c₂,b₁,0] with Case = A or B followed by coordinates of centers A, B, C. For example B,0,0,1,1,1,0 means we are looking at three mutually tangent circles with centers at A(0,0), B(1,1) and C(1,0). In these examples we have included the lines of intersection L_AB(d) and L_AC(d) for each of the values d = D₁ and d = D₂. Also, we use the subscript notation D_1iji on the D-radii to mean the radius obtained when x₁(D_i) = x_j(D_i) for i = 1,2; j = 2,3. The values are given to two decimal places only. Note: Case A 4th tangent circles are shaded purple and Case B circles are shaded red.

• A,0,0,1,1,1,0

  D₁₂₂₂ = 0.14 D(1.37,0.77)
  D₁₁₃₁ = 0.71 D(0,1)

• B,0,0,1,1,1,0

  D₁₂₃₂ = 1.71 D(0,1)
  D₁₁₂₁ = -0.08 D(0.74,0.26)

• Notice the dramatic difference with the large surrounding circle at C(2.8,0) and the externally tangent circle at C(2.9,0)
  B,0,0,1,1,2.8,0

  D₁₂₂₂ = 479.71 D(111.57,-465.45)
  D₁₁₂₁ = -0.11 D(1.05,0.55)

• B,0,0,1,1,2.9,0

  D₁₂₃₂ = -24.00 D(-4.98,24.59)
  D₁₁₂₁ = -0.11 D(1.06,0.56)

• A,0,0,1,1,,4.3,0

  D₁₂₂₂ = 0.21 D(4.34,0.49)
  D₁₁₂₁ = 0.38 D(4.15,-0.65)

• B,0,0,1,1,,4.3,0

  D₁₂₃₂ = -2.03 D(0.17,3.16)
  D₁₁₂₁ = -0.11 D(1.08,0.62)