Simpli fied Solutions of the CLP and CCP Limiting Cases of the Problem of Apollonius via Vector Rotations using Geometric Algebra
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This document presents simplified solutions for the 'clp' and 'ccp' limiting cases of the problem of Apollonius using vector rotations with geometric algebra. The new solutions are more efficient and accessible compared to previous works, stemming from improved angle relationships and better utilization of geometric identities. The document includes detailed discussions of the solution processes, illustrations, and references to prior research.
![ABSTRACT
The new solutions presented herein for the CLP and CCP limiting cases of the
Problem of Apollonius are much shorter and more easily understood than those
provided by the same author in [1]-[2]. These improvements result from (1) a
better selection of angle relationships as a starting point for the solution process;
and (2) better use of GA identities to avoid forming troublesome combinations
of terms within the resulting equations.
3](https://image.slidesharecdn.com/improvedclpandccpsolns-160820123212/85/Simpli-fied-Solutions-of-the-CLP-and-CCP-Limiting-Cases-of-the-Problem-of-Apollonius-via-Vector-Rotations-using-Geometric-Algebra-3-320.jpg)
![1 Introduction
This document shows how the CLP and CCP limiting cases can be solved more
efficiently than in the author’s previous work ( [1] - [2]). Because that work and
[3] discussed the necessary background in detail, the solutions presented here
are somewhat abbreviated.
2 Solution of the CLP Limiting Case
For detailed discussions of the ideas used in this solution, please see [1]. We’ll
show two ways of solving the problem; the second takes advantage of observa-
tions made during the first.
The CLP limiting case reads,
Given a circle C, a line L, and a point P, construct the circles that
are tangent to C and L, and pass through P.
Figure 2.1: The CLP Limiting Case of the Problem of Apollonius: Given a
circle C, a line L, and a point P, construct the circles that are tangent to C and
L, and pass through P.
The problem has two types of solutions:
• Circles that enclose C;
• Circles that do not enclose C.
There are two solution circles of each type. In this document, we’ll treat
only those that do not enclose the given circle.
4](https://image.slidesharecdn.com/improvedclpandccpsolns-160820123212/85/Simpli-fied-Solutions-of-the-CLP-and-CCP-Limiting-Cases-of-the-Problem-of-Apollonius-via-Vector-Rotations-using-Geometric-Algebra-4-320.jpg)
![2.1 The First Solution
Fig. 2.2 shows how we will capture the geometric content of the problem. An
important improvement, compared to the solution technique presented in [1], is
that we will use rotations with respect to the vector from the given point P to
the still-unidentified center point (c2) of the solution circle.
Figure 2.2: Elements used in the first solution of the CLP limiting case.
In deriving our solution, we’ll
use the same symbol —for
example, t —to denote both a
point and the vector to that
point from the origin. We’ll rely
upon context to tell the reader
whether the symbol is being
used to refer to the point, or to
the vector.
We’ll begin our solution by deriving an expression for r2 in terms of ˆt. We´ll
do so by equating two independent expressions for s, then “dotting” both sides
with ˆh, after which we’ll solve for r2:
(r1 + r2)ˆt + r2
ˆh = h + λˆhi
(r1 + r2)ˆt + r2
ˆh · ˆh = h + λˆhi · ˆh
(r1 + r2)ˆt · ˆh + r2
ˆh · ˆh = h · ˆh + λ ˆhi · ˆh
(r1 + r2)ˆt · ˆh + r2 = h + 0;
∴ r2 =
h − r1
ˆt · ˆh
1 + ˆt · ˆh
, and r1 + r2 =
h + r1
1 + ˆt · ˆh
. (2.1)
Next, we equate two expressions for the rotation ei2φ
:
t − p
t − p
c2 − p
c2 − p
=eiφ
t − p
t − p
c2 − p
c2 − p
=eiφ
= −ˆt
c2 − p
c2 − p
=ei2φ
,
from which
[t − p] [c2 − p] [t − p] ˆt = some scalar,
∴ [t − p] [c2 − p] [t − p] ˆt 2 = 0. (2.2)
5](https://image.slidesharecdn.com/improvedclpandccpsolns-160820123212/85/Simpli-fied-Solutions-of-the-CLP-and-CCP-Limiting-Cases-of-the-Problem-of-Apollonius-via-Vector-Rotations-using-Geometric-Algebra-5-320.jpg)
![Using the identity ab ≡ 2a ∧ b + ba, we rewrite 2.2 as
(2 [t − p] ∧ [c2 − p] + [c2 − p] [t − p]) [t − p] ˆt 2 = 0,
(2 [t − p] ∧ [c2 − p]) [t − p] ˆt + [t − p]
2
[c2 − p] ˆt 2 = 0, and
(2 [t − p] ∧ [c2 − p]) [t − p] ˆt 2 + [t − p]
2
[c2 − p] ˆt 2 = 0. (2.3)
Now, we note that
(2 [t − p] ∧ [c2 − p]) [t − p] ˆt 2 = 2 ([t − p] ∧ [c2 − p]) [t − p] · ˆt ,
and [t − p]
2
[c2 − p] ˆt 2 = [t − p]
2
[c2 − p] ∧ ˆt .
Note how the factor p ∧ ˆt
canceled out in Eq. (2.4). That
cancellation suggests an
improvement that we’ll see in
our second solution of the CLP
case.
Because t = r1
ˆt and c2 = (r1 + r2)ˆt, t ∧ c2 = 0. We can expand [t − p]
2
as
r1
2
− 2p · t + p2
. Using all of these ideas, (2.3) becomes (after simplification)
2r2 r1 − p · ˆt p ∧ ˆt + r1
2
− 2p · t + p2
p ∧ ˆt = 0. (2.4)
For p ∧ ˆt = 0, that equation becomes
2r2 r1 − p · ˆt + r1
2
− 2p · t + p2
= 0.
Substituting the expression that we derived for r2 in (2.1), then expanding and
simplifying,
2 ( h + r1) p · ˆt − p2
− r1
2 ˆh · ˆt = 2 h r1 + r1
2
+ p2
.
Finally, we rearrange that result and multiply both sides by r1 h , giving the
equation that we derived in [1]:
2 r1 h + h2
p − p2
− r1
2
h · t = 2h2
r1
2
+ r1 h r1
2
+ p2
. (2.5)
2.2 The Second Solution: Learning From and Building
Upon the First
In Eq. (2.4), we saw how the factor p∧ˆt canceled out. That cancellation suggests
that we might solve the problem more efficiently by expressing rotations with
respect to the unknown vector ˆt, rather than to a vector from P to c2 (Fig.
2.3).
For this new choice of vectors, our equation relating two expressions for
the rotation ei2φ
is:
ˆt
p − t
p − t
=eiφ
ˆt
p − t
p − t
=eiφ
= ˆt
p − c2
p − c2
=ei2φ
,
from which
[p − t] ˆt [p − t] [p − c2] = some scalar,
∴ [p − t] ˆt [p − t] [p − c2] 2 = 0. (2.6)
6](https://image.slidesharecdn.com/improvedclpandccpsolns-160820123212/85/Simpli-fied-Solutions-of-the-CLP-and-CCP-Limiting-Cases-of-the-Problem-of-Apollonius-via-Vector-Rotations-using-Geometric-Algebra-6-320.jpg)
![Figure 2.3: Elements used in the second solution of the CLP Limiting Case:
rotations are now expressed with respect to the unknown vector ˆt, rather than
to a vector from P to c2.
Using the identity ab ≡ 2a ∧ b + ba, we rewrite 2.6 as
2 [p − t] ∧ ˆt + ˆt [p − t] [p − t] [p − c2] 2 = 0, and
2 [p − t] ∧ ˆt [p − t] [p − c2] 2 + [p − t]
2 ˆt [p − c2] 2 = 0. (2.7)
Now, we note that
2 [p − t] ∧ ˆt [p − t] [p − c2] 2 = 2 [p − t] ∧ ˆt [p − t] · [p − c2] ,
and [p − t]
2 ˆt [p − c2] 2 = [p − t]
2 ˆt ∧ [p − c2] .
Because t = r1
ˆt and c2 = (r1 + r2)ˆt, t ∧ c2 = 0. We can expand [p − t]
2
as
p2
− 2p · t + r1
2
. Using all of these ideas, (2.7) becomes (after simplification)
2 ([p − t] · [p − c2]) p ∧ t − p2
− 2p · t + r1
2
p ∧ t = 0.
For p ∧ ˆt = 0, that equation becomes, after expanding [p − t] · [p − c2] and
further simplifications,
p2
− r1
2
− 2p · c2 + 2t · c2 = 0.
Now, recalling that c2 = (r1 + r2)ˆt, we substitute the expression that we
derived for r1 + r2 in (2.1), then expand and simplify to obtain (2.5). This
solution process has been a bit shorter than the first because (2.7) was so easy
to simplify.
3 Solution of the CCP Limiting Case
In this solution, we’ll follow the example of Section 2.2, and use rotations with
respect to the vector ˆt. For detailed discussions of the ideas used in this solution,
please see [2].
7](https://image.slidesharecdn.com/improvedclpandccpsolns-160820123212/85/Simpli-fied-Solutions-of-the-CLP-and-CCP-Limiting-Cases-of-the-Problem-of-Apollonius-via-Vector-Rotations-using-Geometric-Algebra-7-320.jpg)
![we proceed as follows:
(r1 + r3)ˆt − c2 = (r2 + r3) ˆw
(r1 + r3)ˆt − c2
2
= [(r2 + r3) ˆw]
2
(r1 + r3)
2
− 2 (r1 + r3) c2 · ˆt + c2
2
= (r2 + r3)
2
∴ r3 =
c2
2
+ r1
2
− r2
2
− 2r1c2 · ˆt
2 r2 − r1 + c2 · ˆt
,
and r1 + r3 =
c2
2
− (r2 − r1)
2
2 r2 − r1 + c2 · ˆt
. (3.1)
Next, we equate two expressions for the rotation ei2θ
:
p − t
p − t
ˆt
=eiθ
p − t
p − t
ˆt
=eiθ
=
p − c3
p − c3
ˆt
=ei2θ
,
from which
[p − t] ˆt [p − t] [p − c3] = some scalar,
∴ [p − t] ˆt [p − t] [p − c3] 2 = 0. (3.2)
Using the identity ab ≡ 2a ∧ b + ba, we rewrite 3.2 as
2 [p − t] ∧ ˆt + ˆt [p − t] [p − t] [p − c3] 2 = 0, and
2 [p − t] ∧ ˆt [p − t] [p − c3] 2 + [p − t]
2 ˆt [p − c3] 2 = 0. (3.3)
Now, we note that
2 [p − t] ∧ ˆt [p − t] [p − c3] 2 = 2 [p − t] ∧ ˆt [p − t] · [p − c3]
and [p − t]
2 ˆt [p − c3] 2 = [p − t]
2 ˆt ∧ [p − c3] .
Because t = r1
ˆt and c3 = (r1 + r3)ˆt, t ∧ c3 = 0. We can expand [p − t]
2
as
p2
− 2p · t + r1
2
. Using all of these ideas, (3.3) becomes (after simplification)
2 ([p − t] · [p − c3]) p ∧ t − p2
− 2p · t + r1
2
p ∧ t = 0.
For p ∧ ˆt = 0, that equation becomes, after further simplification,
p2
− r1
2
− 2p · c3 + 2t · c3 = 0.
Recalling that c3 = (r1 + r[3])ˆt, and substituting the expression that we derived
for r1 + r3 in (3.1), then expanding and simplifying,
c2
2 − (r2 − r1)
2
p · ˆt − p2
− r1
2
c2 · ˆt = (r2 − r1) p2
− r2r1 + r1c2
2.
Finally, we rearrange that result and multiply both sides by r1, giving
c2
2 − (r2 − r1)
2
p − p2
− r1
2
c2 · t = r1 (r2 − r1) p2
− r2r1 + r1c2
2 . (3.4)
9](https://image.slidesharecdn.com/improvedclpandccpsolns-160820123212/85/Simpli-fied-Solutions-of-the-CLP-and-CCP-Limiting-Cases-of-the-Problem-of-Apollonius-via-Vector-Rotations-using-Geometric-Algebra-9-320.jpg)
![Defining u = c2
2 − (r2 − r1)
2
p − p2
− r1
2
c2, we can transform that result
into
P ˆu (t) =
r1 (r2 − r1) p2
− r2r1 + r1c2
2
u
, (3.5)
where P ˆu (t) is the projection of t upon ˆu . As described in detail in [2], there
are two vectors that fulfill that condition. Labeled ˆt and ˆt in Fig. 3.3, they are
the vectors from the center of C1 to the points of tangency with the two solution
circles shown. Readers are encouraged to derive this same solution using the
Figure 3.3: The solution circles that enclose both of the givens, and that enclose
neither. See text for definitions of u and P ˆu (t).
magenta circle as the starting point.
4 Literature Cited
References
[1] “Solution of the Special Case ‘CLP’ of the Problem of Apollo-
nius via Vector Rotations using Geometric Algebra”. Available at
http://vixra.org/abs/1605.0314.
[2] “The Problem of Apollonius as an Opportunity for Teaching Students to
Use Reflections and Rotations to Solve Geometry Problems via Geometric
(Clifford) Algebra”. Available at http://vixra.org/abs/1605.0233.
[3] J. Smith, “Rotations of Vectors Via Geometric Algebra: Explanation, and
Usage in Solving Classic Geometric ‘Construction’ Problems” (Version of
11 February 2016). Available at http://vixra.org/abs/1605.0232 .
10](https://image.slidesharecdn.com/improvedclpandccpsolns-160820123212/85/Simpli-fied-Solutions-of-the-CLP-and-CCP-Limiting-Cases-of-the-Problem-of-Apollonius-via-Vector-Rotations-using-Geometric-Algebra-10-320.jpg)
Simpli fied Solutions of the CLP and CCP Limiting Cases of the Problem of Apollonius via Vector Rotations using Geometric Algebra
- 1. Simplified Solutions of the CLP and CCP Limiting Cases of the Problem of Apollonius via Vector Rotations using Geometric Algebra 1
- 2. Simplified Solutions of the “CLP” and “CCP” Limiting Cases of the Problem of Apollonius via Vector Rotations using Geometric Algebra Jim Smith QueLaMateNoTeMate.webs.com email: nitac14b@yahoo.com August 19, 2016 Contents 1 Introduction 4 2 Solution of the CLP Limiting Case 4 2.1 The First Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2.2 The Second Solution: Learning From and Building Upon the First 6 3 Solution of the CCP Limiting Case 7 4 Literature Cited 10 2
- 3. ABSTRACT The new solutions presented herein for the CLP and CCP limiting cases of the Problem of Apollonius are much shorter and more easily understood than those provided by the same author in [1]-[2]. These improvements result from (1) a better selection of angle relationships as a starting point for the solution process; and (2) better use of GA identities to avoid forming troublesome combinations of terms within the resulting equations. 3
- 4. 1 Introduction This document shows how the CLP and CCP limiting cases can be solved more efficiently than in the author’s previous work ( [1] - [2]). Because that work and [3] discussed the necessary background in detail, the solutions presented here are somewhat abbreviated. 2 Solution of the CLP Limiting Case For detailed discussions of the ideas used in this solution, please see [1]. We’ll show two ways of solving the problem; the second takes advantage of observa- tions made during the first. The CLP limiting case reads, Given a circle C, a line L, and a point P, construct the circles that are tangent to C and L, and pass through P. Figure 2.1: The CLP Limiting Case of the Problem of Apollonius: Given a circle C, a line L, and a point P, construct the circles that are tangent to C and L, and pass through P. The problem has two types of solutions: • Circles that enclose C; • Circles that do not enclose C. There are two solution circles of each type. In this document, we’ll treat only those that do not enclose the given circle. 4
- 5. 2.1 The First Solution Fig. 2.2 shows how we will capture the geometric content of the problem. An important improvement, compared to the solution technique presented in [1], is that we will use rotations with respect to the vector from the given point P to the still-unidentified center point (c2) of the solution circle. Figure 2.2: Elements used in the first solution of the CLP limiting case. In deriving our solution, we’ll use the same symbol —for example, t —to denote both a point and the vector to that point from the origin. We’ll rely upon context to tell the reader whether the symbol is being used to refer to the point, or to the vector. We’ll begin our solution by deriving an expression for r2 in terms of ˆt. We´ll do so by equating two independent expressions for s, then “dotting” both sides with ˆh, after which we’ll solve for r2: (r1 + r2)ˆt + r2 ˆh = h + λˆhi (r1 + r2)ˆt + r2 ˆh · ˆh = h + λˆhi · ˆh (r1 + r2)ˆt · ˆh + r2 ˆh · ˆh = h · ˆh + λ ˆhi · ˆh (r1 + r2)ˆt · ˆh + r2 = h + 0; ∴ r2 = h − r1 ˆt · ˆh 1 + ˆt · ˆh , and r1 + r2 = h + r1 1 + ˆt · ˆh . (2.1) Next, we equate two expressions for the rotation ei2φ : t − p t − p c2 − p c2 − p =eiφ t − p t − p c2 − p c2 − p =eiφ = −ˆt c2 − p c2 − p =ei2φ , from which [t − p] [c2 − p] [t − p] ˆt = some scalar, ∴ [t − p] [c2 − p] [t − p] ˆt 2 = 0. (2.2) 5
- 6. Using the identity ab ≡ 2a ∧ b + ba, we rewrite 2.2 as (2 [t − p] ∧ [c2 − p] + [c2 − p] [t − p]) [t − p] ˆt 2 = 0, (2 [t − p] ∧ [c2 − p]) [t − p] ˆt + [t − p] 2 [c2 − p] ˆt 2 = 0, and (2 [t − p] ∧ [c2 − p]) [t − p] ˆt 2 + [t − p] 2 [c2 − p] ˆt 2 = 0. (2.3) Now, we note that (2 [t − p] ∧ [c2 − p]) [t − p] ˆt 2 = 2 ([t − p] ∧ [c2 − p]) [t − p] · ˆt , and [t − p] 2 [c2 − p] ˆt 2 = [t − p] 2 [c2 − p] ∧ ˆt . Note how the factor p ∧ ˆt canceled out in Eq. (2.4). That cancellation suggests an improvement that we’ll see in our second solution of the CLP case. Because t = r1 ˆt and c2 = (r1 + r2)ˆt, t ∧ c2 = 0. We can expand [t − p] 2 as r1 2 − 2p · t + p2 . Using all of these ideas, (2.3) becomes (after simplification) 2r2 r1 − p · ˆt p ∧ ˆt + r1 2 − 2p · t + p2 p ∧ ˆt = 0. (2.4) For p ∧ ˆt = 0, that equation becomes 2r2 r1 − p · ˆt + r1 2 − 2p · t + p2 = 0. Substituting the expression that we derived for r2 in (2.1), then expanding and simplifying, 2 ( h + r1) p · ˆt − p2 − r1 2 ˆh · ˆt = 2 h r1 + r1 2 + p2 . Finally, we rearrange that result and multiply both sides by r1 h , giving the equation that we derived in [1]: 2 r1 h + h2 p − p2 − r1 2 h · t = 2h2 r1 2 + r1 h r1 2 + p2 . (2.5) 2.2 The Second Solution: Learning From and Building Upon the First In Eq. (2.4), we saw how the factor p∧ˆt canceled out. That cancellation suggests that we might solve the problem more efficiently by expressing rotations with respect to the unknown vector ˆt, rather than to a vector from P to c2 (Fig. 2.3). For this new choice of vectors, our equation relating two expressions for the rotation ei2φ is: ˆt p − t p − t =eiφ ˆt p − t p − t =eiφ = ˆt p − c2 p − c2 =ei2φ , from which [p − t] ˆt [p − t] [p − c2] = some scalar, ∴ [p − t] ˆt [p − t] [p − c2] 2 = 0. (2.6) 6
- 7. Figure 2.3: Elements used in the second solution of the CLP Limiting Case: rotations are now expressed with respect to the unknown vector ˆt, rather than to a vector from P to c2. Using the identity ab ≡ 2a ∧ b + ba, we rewrite 2.6 as 2 [p − t] ∧ ˆt + ˆt [p − t] [p − t] [p − c2] 2 = 0, and 2 [p − t] ∧ ˆt [p − t] [p − c2] 2 + [p − t] 2 ˆt [p − c2] 2 = 0. (2.7) Now, we note that 2 [p − t] ∧ ˆt [p − t] [p − c2] 2 = 2 [p − t] ∧ ˆt [p − t] · [p − c2] , and [p − t] 2 ˆt [p − c2] 2 = [p − t] 2 ˆt ∧ [p − c2] . Because t = r1 ˆt and c2 = (r1 + r2)ˆt, t ∧ c2 = 0. We can expand [p − t] 2 as p2 − 2p · t + r1 2 . Using all of these ideas, (2.7) becomes (after simplification) 2 ([p − t] · [p − c2]) p ∧ t − p2 − 2p · t + r1 2 p ∧ t = 0. For p ∧ ˆt = 0, that equation becomes, after expanding [p − t] · [p − c2] and further simplifications, p2 − r1 2 − 2p · c2 + 2t · c2 = 0. Now, recalling that c2 = (r1 + r2)ˆt, we substitute the expression that we derived for r1 + r2 in (2.1), then expand and simplify to obtain (2.5). This solution process has been a bit shorter than the first because (2.7) was so easy to simplify. 3 Solution of the CCP Limiting Case In this solution, we’ll follow the example of Section 2.2, and use rotations with respect to the vector ˆt. For detailed discussions of the ideas used in this solution, please see [2]. 7
- 8. The CCP limiting case reads, “Given two circles and a point P, all coplanar, construct the circles that pass through P and are tangent, simultaneously, to the given circles.” (Fig. 3.1). Figure 3.1: The CCP limiting case of the Problem of Apollonius: “Given two circles (C1, C2) and a point P, all coplanar, construct the circles that pass through P and are tangent, simultaneously, to the given circles.” We´ll derive the solution for the solution circles that enclose either both of the given ones, or neither. Fig. 3.2 shows how we’ll capture the geometric content. As in the CLP case, we can find both solution circles of this type by analyzing the diagram for just one of them. Figure 3.2: Elements used in the solution of the CCP limiting case. We begin the solution by deriving an expression for r3 in terms of ˆt and the given quantities. From two independent equations for c3, (r1 + r3)ˆt = c3 = c2 + (r2 + r3) ˆw, 8
- 9. we proceed as follows: (r1 + r3)ˆt − c2 = (r2 + r3) ˆw (r1 + r3)ˆt − c2 2 = [(r2 + r3) ˆw] 2 (r1 + r3) 2 − 2 (r1 + r3) c2 · ˆt + c2 2 = (r2 + r3) 2 ∴ r3 = c2 2 + r1 2 − r2 2 − 2r1c2 · ˆt 2 r2 − r1 + c2 · ˆt , and r1 + r3 = c2 2 − (r2 − r1) 2 2 r2 − r1 + c2 · ˆt . (3.1) Next, we equate two expressions for the rotation ei2θ : p − t p − t ˆt =eiθ p − t p − t ˆt =eiθ = p − c3 p − c3 ˆt =ei2θ , from which [p − t] ˆt [p − t] [p − c3] = some scalar, ∴ [p − t] ˆt [p − t] [p − c3] 2 = 0. (3.2) Using the identity ab ≡ 2a ∧ b + ba, we rewrite 3.2 as 2 [p − t] ∧ ˆt + ˆt [p − t] [p − t] [p − c3] 2 = 0, and 2 [p − t] ∧ ˆt [p − t] [p − c3] 2 + [p − t] 2 ˆt [p − c3] 2 = 0. (3.3) Now, we note that 2 [p − t] ∧ ˆt [p − t] [p − c3] 2 = 2 [p − t] ∧ ˆt [p − t] · [p − c3] and [p − t] 2 ˆt [p − c3] 2 = [p − t] 2 ˆt ∧ [p − c3] . Because t = r1 ˆt and c3 = (r1 + r3)ˆt, t ∧ c3 = 0. We can expand [p − t] 2 as p2 − 2p · t + r1 2 . Using all of these ideas, (3.3) becomes (after simplification) 2 ([p − t] · [p − c3]) p ∧ t − p2 − 2p · t + r1 2 p ∧ t = 0. For p ∧ ˆt = 0, that equation becomes, after further simplification, p2 − r1 2 − 2p · c3 + 2t · c3 = 0. Recalling that c3 = (r1 + r[3])ˆt, and substituting the expression that we derived for r1 + r3 in (3.1), then expanding and simplifying, c2 2 − (r2 − r1) 2 p · ˆt − p2 − r1 2 c2 · ˆt = (r2 − r1) p2 − r2r1 + r1c2 2. Finally, we rearrange that result and multiply both sides by r1, giving c2 2 − (r2 − r1) 2 p − p2 − r1 2 c2 · t = r1 (r2 − r1) p2 − r2r1 + r1c2 2 . (3.4) 9
- 10. Defining u = c2 2 − (r2 − r1) 2 p − p2 − r1 2 c2, we can transform that result into P ˆu (t) = r1 (r2 − r1) p2 − r2r1 + r1c2 2 u , (3.5) where P ˆu (t) is the projection of t upon ˆu . As described in detail in [2], there are two vectors that fulfill that condition. Labeled ˆt and ˆt in Fig. 3.3, they are the vectors from the center of C1 to the points of tangency with the two solution circles shown. Readers are encouraged to derive this same solution using the Figure 3.3: The solution circles that enclose both of the givens, and that enclose neither. See text for definitions of u and P ˆu (t). magenta circle as the starting point. 4 Literature Cited References [1] “Solution of the Special Case ‘CLP’ of the Problem of Apollo- nius via Vector Rotations using Geometric Algebra”. Available at http://vixra.org/abs/1605.0314. [2] “The Problem of Apollonius as an Opportunity for Teaching Students to Use Reflections and Rotations to Solve Geometry Problems via Geometric (Clifford) Algebra”. Available at http://vixra.org/abs/1605.0233. [3] J. Smith, “Rotations of Vectors Via Geometric Algebra: Explanation, and Usage in Solving Classic Geometric ‘Construction’ Problems” (Version of 11 February 2016). Available at http://vixra.org/abs/1605.0232 . 10