Calculation of External straight lines tangent to 2 circles

The goal is here to calculate the coordinates of the points connecting the external straight lines tangent to 2 circles (see figures herebellow).

Two cases have to be considered:

  1. The general case where R_1 \neq R_2,
  2. The case where R_1 = R_2.
circles_tangent_steps_final
circles_idem_final
(circles where R_1 \ne R_2) (circles where R_1 = R_2)

Remark: As far as possible, the equations will be simplified.

 

 

1- Case where R_1 \neq R_2

1-1 The principle

By beginning by the general case, i.e. the one where the 2 radii are different (and non equal to zero of course), some particularities will be highlighted…

 

As we did in the post on the fillets calculation (see post fillet calculation in 2D for more information), the tangent points are the intersecting points of the circles:

  • \mathcal{C'}_1 of centre A' where A' is the middle of \left[ AC \right] and of radius {R'}_1 = \frac{1}{2} || AC ||,
  • \mathcal{C'}_2 of centre B' where B'is the middle of  \left[ BC \right] and of radius {R'}_2 = \frac{1}{2} || BC ||.

 

circles_tangent_principe
(Principle of the calculation of the tangent points)

 

 

1-2) Calculation of the centre dilatation C 

Since R_1 \neq R_2, a dilatation centre C exists that transforms \mathcal{C}_1 in \mathcal{C}_2 (see figure hereafter).

 

circles_tangent_steps_1
(case whereR_1 \neq R_2)

Let the point C with the coordinates \left( \begin{array}{c} x_C\\ y_C \end{array} \right)

 

From Thalès’ theorem, we can write that \frac{R_2}{R_1}=\frac{[ BC ]}{[ AC ]} = \frac{[ CB ]}{[ CA ]}.

 

Then R_2 \cdot || CA || = R_1 \cdot ||CB ||, or even  R_2 \left( \begin{array}{c} x_A - x_C \\ y_A - y_C \end{array} \right) = R_1 \left( \begin{array}{c} x_B - x_C \\ y_B - y_C \end{array} \right)

 

So the system becomes
 \left\lbrace \begin{array}{l} R_2 \left( x_A - x_C \right) = R_1 \left( x_B - x_C \right) \\ R_2 \left( y_A - y_C \right) = R_1 \left( y_B - y_C \right) \end{array} \right.

 

After calculations, the point C expresses as:

(1)   \begin{equation*} \fbox{$ \left\lbrace \begin{array}{l} x_C = \left( \frac{R_2}{R_2 - R_1} \right) x_A - \left( \frac{R_1}{R_2 - R_1} \right) x_B \\ y_C = \left( \frac{R_2}{R_2 - R_1} \right) y_A - \left( \frac{R_1}{R_2 - R_1} \right) y_B \end{array} \right.  $} \end{equation*}

 

 

1-3) Calculation of the tangent points to \mathcal{C}_1 

As told previously, these are the intersecting points between the circles \mathcal{C}_1 and \mathcal{C'}_1 so that:

 \left\lbrace \begin{array}{rl} \mathcal{C}_1 : & \left( x - x_A \right)^2 + \left( y - y_A \right)^2 = R_1^2 \\ \mathcal{C'}_1: & \left( x - x_{A'} \right)^2 + \left( y - y_{A'} \right)^2 = {R'}_1^2 \\ \end{array} \right.

(Remind that A' is the middle of [AC]).

 

By substracting the equation of \mathcal{C'}_1 to the one of \mathcal{C}_1, it leads to the new equation:

 y = - \left( \frac{x_{A'} - x_A}{y_{A'} - y_A} \right)x + \left[ \frac{\left( R_1^2 - {R'}_1^2 \right) - \left( x_A^2 - x_{A'}^2 \right) - \left( y_A^2 - y_{A'}^2 \right)}{2 \left( y_{A'} - y_A \right)} \right]

 

The later one is rewritten y = \alpha_1 x + N_1 with \left\lbrace \begin{array}{l} \alpha_1 = \left( \frac{x_{A'} - x_A}{y_{A'} - y_A} \right) \\ N_1 = \left[ \frac{\left( R_1^2 - {R'}_1^2 \right) - \left( x_A^2 - x_{A'}^2 \right) - \left( y_A^2 - y_{A'}^2 \right)}{2 \left( y_{A'} - y_A \right)} \right] \end{array} \right.

 

Since A' is the middle of [AC], then\left\lbrace \begin{array}{l} x_{A'} = \frac{1}{2} \left( x_A + x_c \right) \rightarrow \left( x_{A'} - x_A \right) = \frac{x_C - x_A}{2} \\ y_{A'} = \frac{1}{2} \left( y_A + y_c \right) \rightarrow \left( y_{A'} - y_A \right) = \frac{y_C - y_A}{2} \end{array} \right.

 

Eq (1) is reinjected into the previous equations and we get:
\left\lbrace \begin{array}{l} \left( x_{A'} - x_A \right) = \frac{1}{2} \left( \frac{R_1}{R2 - R_1} \right) \left( x_A - x_B \right)\\ \left( y_{A'} - y_A \right) = \frac{1}{2} \left( \frac{R_1}{R2 - R_1} \right) \left( y_A - y_B \right) \end{array} \right.

 

And then:

(2)   \begin{equation*} \fbox{$ \alpha_1 = \left( \frac{x_{A'} - x_A}{y_{A'} - y_A} \right) = \left( \frac{x_A - x_B}{y_A - y_B} \right)  $} \end{equation*}

 

Since:
\left\lbrace \begin{array}{l} {R'}_1 = \frac{1}{2}\left[  AC \right] \text{ then }{R'}_1^2 = \frac{1}{4} \left[ AC \right]^2 = \frac{1}{4} \left( x_A - x_C \right)^2 + \frac{1}{4} \left( y_A - y_C \right)^2 \\ \text{but } \left( x_{A'} - x_A \right) = \frac{x_C - x_A}{2} \\ \text{and }\left( y_{A'} - y_A \right) = \frac{y_C - y_A}{2} \\ \text{then }{R'}_1^2 \text{ is rewritten as } {R'}_1^2 = \left( x_A - x_{A'}\right)^2 + \left( y_A - y_{A'}\right)^2 \\ \text{by reinjection into }N_1\text{, it leads to } N_1 = \left[ \frac{R_1^2 - 2 x_A^2 - 2 y_A^2 + 2 x_A x_{A'} + 2 y_A y_{A'}}{2 \left( y_{A'} - y_A \right)} \right] \end{array} \right.

 

From eq. (1), the “prime” terms are recalculated:
\left\lbrace \begin{array}{l} x_{A'} = \left( \frac{x_A + x_C}{2} \right) = \frac{1}{2} \left( \frac{2 R_2 x_A - R_1 x_A - R_1 x_B}{R_2 - R_1} \right) \\ y_{A'} = \left( \frac{y_A + y_C}{2} \right) = \frac{1}{2} \left( \frac{2 R_2 y_A - R_1 y_A - R_1 y_B}{R_2 - R_1} \right) \\ 2 x_A x_{A'} = x_A \left[ \left( \frac{2 R_2 - R_1}{R_2 - R_1} \right) x_A - \left( \frac{R_1}{R_2 - R_1} \right)x_B \right] \\ 2 y_A y_{A'} = y_A \left[ \left( \frac{2 R_2 - R_1}{R_2 - R_1} \right) y_A - \left( \frac{R_1}{R_2 - R_1} \right)y_B \right] \end{array} \right.

 

Using the previous equations, including them into N_1 we finally have:

(3)   \begin{equation*} \fbox{$ N_1 = \left( \frac{R_1 R_2 - R_1^2 + x_A^2 + y_A^2 - x_A x_B - y_A y_B}{y_A - y_B} \right)  $} \end{equation*}

 

\alpha_1 equation and N_1 one are reinjected into the equation of \mathcal{C}_1 to lead to a 2nd degree equation:
A_1 x^2 + B_1 x + C_1 = 0 with  \left\lbrace \begin{array}{l} A_1 = \left( 1 + \alpha_1 \right)^2 \\  B_1 = 2 \alpha_1 \left( y_A - N_1 \right) - 2 x_A \\ C_1 = \left( x_A^2 + y_A^2 \right)  + N_1 \left( N_1 - 2 y_A \right) - R_1^2 \end{array} \right.

 

Therefore

(4)   \begin{equation*} \fbox{$\left\lbrace \begin{array}{l} \Delta_1 = B_1^2 + 4 A_1 C_1 \\  x_{1i} = \frac{- B_1 + k_{1i} \sqrt{\Delta_1}}{2 A_1} \text{ with } k_{1i} = \pm 1 \text{ and } i=1,2\\ y_{1i} = - \alpha_1 x_{1i} + N1 \text{ with }i=1,2 \end{array} \right.  $} \end{equation*}

 

 

 

1-4) Calculation of the tangent points to \mathcal{C}_2

We proceed to the same manner as we did previously, and in order to highlight the current document, only the main results are presented:

(5)   \begin{equation*} \fbox{$ \alpha_2 = \left( \frac{x_B - x_B}{y_B - y_A} \right) = \left( \frac{x_A - x_B}{y_A - y_B} \right) = \alpha_1  $} \end{equation*}

 

and

(6)   \begin{equation*} \fbox{$ N_2 = \left( \frac{R_2^2 - R_1 R_2 + x_A x_B + y_A y_B - \left( x_B^2 + y_B^2 \right) }{y_A - y_B} \right)  $} \end{equation*}

 

 

By reinjection into \mathcal{C}_2 equation to have: \left\lbrace \begin{array}{l} A_2 = \left( 1 + \alpha_2 \right)^2 \\  B_2 = 2 \alpha_2 \left( y_B - N_2 \right) - 2 x_B \\ C_2 = \left( x_B^2 + y_B^2 \right) + N_2 \left( N_2 - 2 y_B \right) - R_2^2 \end{array} \right.

 

 

And then: 

(7)   \begin{equation*} \fbox{$\left\lbrace \begin{array}{l} \Delta_2 = B_2^2 + 4 A_2 C_2 \\  x_{2i} = \frac{- B_2 + k_{2i} \sqrt{\Delta_2}}{2 A_2} \text{ with } k_{2i} = \pm 1 \text{ and } i=1,2\\ y_{2i} = - \alpha_2 x_{2i} + N2 \text{ with }i=1,2 \end{array} \right.  $} \end{equation*}

 

 

Discussions on the values of k_{1i} and k_{2i}:

We know that the angles \widehat{A T_{1i} C} and \widehat{B T_{2i} C} (with i=1,2) are right; then the \cos of the angles is equal to zero.

 

The Algorithm could be:

for k1 = 1
if cos(angle(T_i A C)) = 0 +/- precision
	then ok
	else k1 = -1
end if 

 

In another way, we can notice that the sign of k_{11} (as well as the one of k_{21})  is the opposite of the sign of the slope m (remind: m is the slope of the straight line (AB)) :

– if m > 0 then \left\lbrace \begin{array}{l} T_{i1} \rightarrow k_{i1} = -1 = -\text{sign}(m) \text{ with } i =1,2 \\  T_{i2} \rightarrow k_{i2} = +1 = +\text{sign}(m) = -k_{11} \text{ with } i =1,2 \end{array} \right.

– if m < 0 then \left\lbrace \begin{array}{l} T_{i1} \rightarrow k_{i1} = +1 = +\text{sign}(m) \text{ with } k =1,2\\  T_{i2} \rightarrow k_{i2} = -1 = -\text{sign}(m) = -k_{11} \text{ with } k =1,2 \end{array} \right.

 

Remarks:

  • remind that the case where m = 0 (i.e. y_A = y_B) is a case apart that will be specifically treated,
  • T_{11} and T_{21} are the upper points of respectively the \mathcal{C}_1 circle and the \mathcal{C}_2 one;
  • and T_{11},  T_{21} are le lower points.

 

From the previous remarks, the equations (4)  and (7) are rewritten:

T_{11} =  \left( \begin{array}{c} x_{11} = \frac{- B_1 + k_1 \sqrt{\Delta_1}}{2 A_1} \\ y_{11} = - \alpha_1 x_{11} + N1  \end{array} \right)T_{12} =  \left( \begin{array}{c} x_{12} = \frac{- B_1 - k_1 \sqrt{\Delta_1}}{2 A_1} \\ y_{12} = - \alpha_1 x_{12} + N1  \end{array} \right) ;

T_{21} =  \left( \begin{array}{c} x_{21} = \frac{- B_2 + k_1 \sqrt{\Delta_2}}{2 A_2} \\ y_{21} = - \alpha_2 x_{21} + N2  \end{array} \right)T_{11} =  \left( \begin{array}{c} x_{22} = \frac{- B_2 - k_1 \sqrt{\Delta_2}}{2 A_2} \\ y_{22} = - \alpha_2 x_{22} + N2  \end{array} \right) ;

(with k_1 = - \text{sign}(m)).

 

 

 

1-5) Particular Cases

The equations  (2), (3), (5) and (6) show it is necessary to treat the cases where x_A = x_B (ditto for y_A = y_B) i.e. when the denominator(s) become(s) equal to zero.

 

a) case where y_A = y_B

cas_general2
(case where y_A = y_B)

 

The previous calculations are done from the beginning by noticing that y_A = y_B = y_C; the system is simplified to:

For the circle \mathcal{C}_1

(8)   \begin{equation*} \fbox{$ \left\lbrace \begin{array}{l} x_{11} = x_{12} = x_1 = \frac{R_1 \left(R_2 - R_1 \right) + x_A^2 - x_A x_B}{x_A - x_B}  \\ y_{1i} = y_A + k_{1i} R_1 \sqrt{1 - \left( \frac{R_2 - R_1}{x_A - x_B} \right)^2}  \text{ with } k_{1i} = \pm 1 \end{array} \right.  $} \end{equation*}

 

For the circle \mathcal{C}_2

(9)   \begin{equation*} \fbox{$ \left\lbrace \begin{array}{l} x_{21} = x_{22} = x_2 = \frac{R_2 \left(R_2 - R_1 \right) + x_A x_B - x_B^2}{x_A - x_B}  \\ y_{2i} = y_B + k_{2i} R_2 \sqrt{1 - \left( \frac{R_2 - R_1}{x_A - x_B} \right)^2}   \text{ with } k_{2i} = \pm 1 \end{array} \right.  $} \end{equation*}

 

 

Discussions on the values of k_{1i} and k_{2i}:
Basically, since x_{11} = x_{12} = x_1 and x_{21} = x_{22} = x_2, then:

  • if k_{11}=+1 and k_{21}=+1, we’re on the points y_{11} and y_{21} that have been chosen as the upper points,
  • and k_{21}=-1=-k_{11} and k_{22}=-1=-k_{21}, we’re on y_{12} et y_{22} that are the lower ones.

 

By considering a single variable k_1, the equations are simplified to:

T_{11} =  \left( \begin{array}{c} \frac{R_1 \left(R_2 - R_1 \right) + x_A^2 - x_A x_B}{x_A - x_B} \\ y_A + k_1 R_1 \sqrt{1 - \left( \frac{R_2 - R_1}{x_A - x_B} \right)^2} \end{array} \right)  ;  T_{12} =  \left( \begin{array}{c} \frac{R_1 \left(R_2 - R_1 \right) + x_A^2 - x_A x_B}{x_A - x_B} \\ y_A - k_1 R_1 \sqrt{1 - \left( \frac{R_2 - R_1}{x_A - x_B} \right)^2} \end{array} \right)  ;  

T_{21} =  \left( \begin{array}{c} \frac{R_2 \left(R_2 - R_1 \right) + x_A x_B - x_B^2}{x_A - x_B}  \\ y_B + k_1 R_2 \sqrt{1 - \left( \frac{R_2 - R_1}{x_A - x_B} \right)^2}  \end{array} \right)  ;  T_{22} =  \left( \begin{array}{c} \frac{R_2 \left(R_2 - R_1 \right) + x_A x_B - x_B^2}{x_A - x_B}  \\ y_B - k_1 R_2 \sqrt{1 - \left( \frac{R_2 - R_1}{x_A - x_B} \right)^2}  \end{array} \right)  ;  

 

 

b) case where x_A = x_B

cas_general4 cas_general3
(case where x_A = x_B)

 

The previous results are re-used by noticing that x_A = x_B = x_C, and by circular permutation, the system becomes:

For the circle \mathcal{C}_1

(10)   \begin{equation*} \fbox{$ \left\lbrace \begin{array}{l} x_{1i} = x_A + k_{1i} R_1 \sqrt{1 - \left( \frac{R_2 - R_1}{y_A - y_B} \right)^2}  \text{ with } k_{1i} = \pm 1 \\ y_1 = \frac{R_1 \left(R_2 - R_1 \right) + y_A^2 - y_A y_B}{y_A - y_B} \end{array} \right.  $} \end{equation*}

 

For the circle \mathcal{C}_2

(11)   \begin{equation*} \fbox{$ \left\lbrace \begin{array}{l} x_{2i} = x_B + k_{2i} R_2 \sqrt{1 - \left( \frac{R_2 - R_1}{y_A - y_B} \right)^2}  \text{ with } k_{2i} = \pm 1 \\ y_2 = \frac{R_2 \left(R_2 - R_1 \right) + y_A y_B - y_B^2}{y_A - y_B} \end{array} \right.  $} \end{equation*}

 

 

Discussions on the values of k_{1i} and k_{2i}:

Again by circular permutation:

  • if k_{11}=+1 and k_{21}=+1, we have the points y_{11} et y_{21} which we’re the upper points and they become now the ones on the right,
  • and si k_{12}=-1=-k_{11} and k_{22}=-1=-k_{21},  the points y_{12} et y_{22} move on the left.

 

By considering a single variable k_1, the equations are simplified to:

T_{11} =  \left( \begin{array}{c} x_A + k_1 R_1 \sqrt{1 - \left( \frac{R_2 - R_1}{y_A - y_B} \right)^2} \\ \frac{R_1 \left(R_2 - R_1 \right) + y_A^2 - y_A y_B}{y_A - y_B} \end{array} \right)  ;  T_{12} =  \left( \begin{array}{c} x_B - k_1 R_1 \sqrt{1 - \left( \frac{R_2 - R_1}{y_A - y_B} \right)^2} \\ \frac{R_1 \left(R_2 - R_1 \right) + y_A^2 - y_A y_B}{y_A - y_B} \end{array} \right)  ;  

T_{21} =  \left( \begin{array}{c} x_B + k_1 R_2 \sqrt{1 - \left( \frac{R_2 - R_1}{y_A - y_B} \right)^2}  \\ \frac{R_2 \left(R_2 - R_1 \right) + y_A y_B - y_B^2}{y_A - y_B} \end{array} \right)  ;  T_{22} =  \left( \begin{array}{c} x_B - k_1 R_2 \sqrt{1 - \left( \frac{R_2 - R_1}{y_A - y_B} \right)^2}  \\ \frac{R_2 \left(R_2 - R_1 \right) + y_A y_B - y_B^2}{y_A - y_B} \end{array} \right)  ;  

 

 

2) Particular case R_1R_2

R1_eq_R2_cas1 R1_eq_R2_cas2
(case where x_A \neq x_B and y_A \neq y_B) (case where x_A \neq x_B and y_A = y_B)
R1_eq_R2_cas3  
(case where x_A =x_B and y_A \neq y_B)  

 

No dilatation since the tangent lines are parallel to (AB). Then 2 different strategies are possible:

  • the calculation of the equations of the tangents using derivatives,
  • the calculation of the distance d from a point of the tangent to the straight line (AB), knowing that d = \frac{ax + by + c}{\sqrt{a^2 + b^2}} where ax + by + c = 0 is the equation of  (AB).

Only the first approach using derivatives has been used here (case 1.).

 

 

2-1 General case

Remind first that, since the 2 straight lines are parallel, they have the same slope m that can be expressed as:

(12)   \begin{equation*} \fbox{$ m = \frac{\Delta y}{\Delta x}= \left( \frac{y_B - y_A}{x_B - x_A} \right)  $} \end{equation*}

Remind as well that R_1 = R_2 = R; the circles equations are consequently:\left\lbrace \begin{array}{rl} \mathcal{C}_1 : & \left( x - x_A \right)^2 + \left( y - y_A \right)^2 = R^2 \\ \mathcal{C}_2: & \left( x - x_B \right)^2 + \left( y - y_B \right)^2 = R^2 \\ \end{array} \right.

(where f'(x) et g'(x) are the derivative functions of f(x) and g(x)).

 

The derivatives are rewritten:
\left\lbrace \begin{array}{rl} \mathcal{C}_1 : & f(x) = y = y_A + k_{1i} \sqrt{R^2 - \left( x - x_A \right)^2} \\ \mathcal{C}_2: & g(x) = y = y_B + k_{2i} \sqrt{R^2 - \left( x - x_B \right)^2} \end{array} \right.
with k_{1i} = \pm 1 and  k_{2i} = \pm 1.

 

Do not forget the general expressions of the tangent equation on the points T_{1i} and T_{2i} :

\left\lbrace \begin{array}{l} y = f' \left( x_{T_{1i}}\right) \left( x - x_{T_{1i}} \right) + f \left( x_{T_{1i}} \right) \\ \text{ with }f' \left( x \right) = k_{1i} \frac{x_A - x}{\sqrt{R^2 - \left( x - x_A \right)^2}} \text{ where } k_{1i} = \pm 1 \end{array} \right.

 

\left\lbrace \begin{array}{l} y = f' \left( x_{T_{1i}}\right) \left( x - x_{T_{1i}} \right) + f \left( x_{T_{1i}} \right) \\ \text{ with }g' \left( x \right) = k_{2i} \frac{x_B - x}{\sqrt{R^2 - \left( x - x_B \right)^2}} \text{ where }k_{2i} = \pm 1 \end{array} \right.

 

And T_{1i}, f' \left( x_{T_{1i}}\right) = m,

i.e. f' \left( x_{T_{1i}} \right) = k_{1i} \frac{x_A - x_{T_{1i}}}{\sqrt{R^2 - \left( x_{T_{1i}} - x_A \right)^2}} = m

 

 

Then after developping 

(13)   \begin{equation*} \fbox{$ x_{T_{1i}} = x_A + k_{1i} \frac{m R}{\sqrt{1 + m^2}}  $} \end{equation*}

 

After reinjection of eq(13) into the circle equation (and after noticing during the development that k_{1i}^2 = 1), then:

(14)   \begin{equation*} \fbox{$ y_{T_{1i}} = y_A + k_{2i} R \sqrt{1 + \frac{m^2}{1+m^2}}  $} \end{equation*}

 

 

We proceed in the same manner for T_{2i} with g' \left( x_{T_{1i}}\right) :

(15)   \begin{equation*} \fbox{$ x_{T_{2i}} = x_B + k_{1i} \frac{m R}{\sqrt{1 + m^2}}  $} \end{equation*}

 

and

 

(16)   \begin{equation*} \fbox{$ y_{T_{2i}} = y_B + k_{2i} R \sqrt{1 + \frac{m^2}{1+m^2}}  $} \end{equation*}

 

 

 

Discussions on the values of k_{1i} and k_{2i}:

For k_1 : As for the general case, the sign of k_{1i} and k_{2i} depends on the one of the slope m:

 

– if m > 0 then \left\lbrace \begin{array}{l} T_{i1} \rightarrow k1_{i1} = -1 = -\text{sign}(m) \text{ with } i =1,2 \\  T_{i2} \rightarrow k1_{i2} = +1 = +\text{sign}(m) = -k_{11} \text{ with } i =1,2 \end{array} \right.

 

– if m < 0 then \left\lbrace \begin{array}{l} T_{i1} \rightarrow k1_{i1} = +1 = +\text{sign}(m) \text{ with } i =1,2\\  T_{i2} \rightarrow k1_{i2} = -1 = -\text{sign}(m) = -k_{11} \text{ with } i =1,2 \end{array} \right.

 

 

For k_2 : we can note that the sign is the opposite of the k_1 one, that allows to rewrite the previous equations using a single variable k_1 where k_1 = - \text{sign}(m):

T_{11} =  \left( \begin{array}{c} x_A + k_1 \frac{m R}{\sqrt{1 + m^2}} \\ y_A - k_1 R \sqrt{1 + \frac{m^2}{1+m^2}} \end{array} \right)  ;  T_{12} =  \left( \begin{array}{c} x_A - k_1 \frac{m R}{\sqrt{1 + m^2}} \\ y_A + k_1 R \sqrt{1 + \frac{m^2}{1+m^2}} \end{array} \right)  ;  

 

T_{21} =  \left( \begin{array}{c} x_B + k_1 \frac{m R}{\sqrt{1 + m^2}} \\ y_B - k_1 R \sqrt{1 + \frac{m^2}{1+m^2}} \end{array} \right)  ;  T_{22} =  \left( \begin{array}{c} x_B - k_1 \frac{m R}{\sqrt{1 + m^2}} \\ y_B + k_1 R \sqrt{1 + \frac{m^2}{1+m^2}} \end{array} \right)

 

 

2-2 Case where y_A = y_B

The equations remain quite simple and can be expressed as:

T_{11} =  \left( \begin{array}{c} x_A \\ y_A + R  \end{array} \right)  ;  T_{12} =  \left( \begin{array}{c} x_A  \\ y_A -  R  \end{array} \right)  ;  T_{21} =  \left( \begin{array}{c} x_B \\ y_B + R  \end{array} \right)  ;  T_{22} =  \left( \begin{array}{c} x_B  \\ y_B - R  \end{array} \right)

Please note that the previous equations remain valid (see section 2-1) since m = 0 in that particular case.

 

 

2-3 Case where x_A = x_B

Unlike the previous particular case, the general equations cannot be applied as it stands since (x_A - x_B) is at the denominator; nevertheless the system remains quite simple and it can be expressed as:

T_{11} =  \left( \begin{array}{c} x_A + R\\ y_A  \end{array} \right)  ;  T_{12} =  \left( \begin{array}{c} x_A  - R\\ y_A   \end{array} \right)  ;  T_{21} =  \left( \begin{array}{c} x_B + R\\ y_B  \end{array} \right)  ;  T_{22} =  \left( \begin{array}{c} x_B  - R\\ y_B  \end{array} \right)

 

 

 

3) Checks

 cas_general1  cas_general2
 cas_general4 R1_eq_R2_cas2
R1_eq_R2_cas1 R1_eq_R2_cas3

An Excel® file (see check_circles_tangentes – compressed with 7z) presents the checks (9 cases) – Geogebra files have been inserted into.

 

 

 

4) Scilab programs

In coming

This work