Demonstração construtiva
Author: Atractor
Date: 25 novembro 2014

$\small\mbox{Move the point or click the button to follow the successive steps.}$

$\small\mbox{Consider two disjoint }  \textcolor{Blue}{circles} \mbox{, with centers A and B and radii } r_{1} \small\mbox{ and }$
$\small  r_{2} \mbox{ (all manipulable).} $

$\small\mbox{Let r be the straight line that links the centers of the two clrcles.}$

$\small\mbox{Let E and F be intersection points of line r with the two } \textcolor{blue}{circles},$
$\small\mbox{(both in the same side with respect to the corresponding centers).}$

$\small\mbox{We shall use the inversion in }  \definecolor{rosa}{rgb}{1,0.6,0.8} \textcolor{rosa}{circle\;C_{1}} \mbox{ 
of center E that goes through F.}$

$\small\mbox{What are the } \definecolor{ciano}{rgb}{0.2,1,0.8} \textcolor{ciano}{inverses} \mbox{ of the } \textcolor{Blue}{\mbox{original circles}}, \mbox{ with respect to } \textcolor{rosa}{C_{1}}?$ 
$\small\mbox{One passes through the inversion center E thus its inverse is a }\textcolor{ciano}{\mbox{straight }} $ 
$ \textcolor{ciano}{\mbox{line (t)}} \small\mbox{ orthogonal to EF.  The inverse of the other } \textcolor{ciano}{circle} \mbox{ is a circle}$
$\small\mbox{tangent to itself in F. }$

$\small\mbox{We denote by G the center of the } \textcolor{ciano}{\mbox{light blue circle }} \mbox{ and by L an}$
$\small\mbox{intersection point of the } \textcolor{ciano}{circle} \mbox{ with the parallel line to }\textcolor{ciano}{t} \mbox{ in G}.$
 

$\small\mbox{Let M be the intersection point of } \textcolor{ciano}{line\;t} \mbox{ with the parallel line to AB } \ \mbox {in L.}$

$\small\mbox{From the properties of } \textcolor{castclaro}{brown \;circle} \mbox{ emphasized in the previous step, }$ 
$\small \mbox{it follows analogous properties for its } \definecolor{castescuro}{rgb}
{0.8,0.4,0} \textcolor{castescuro}{\mbox{inverse: the new circle (darker)}} $
$\small\mbox{is orthogonal to the inverses of the } \textcolor{ciano}{\mbox{light blue objects}}  \mbox{, that is, to the}$
$\small \mbox{two } \textcolor{Blue}{\mbox{initial circles}} \mbox{, and it also intersects line r (at two points).}

$\small \mbox{Since the } \textcolor{castescuro}{\mbox{brown circle}} \mbox{ passes through the inversion center O, its inverse }$
$\small \mbox{is a } \textcolor{castclaro}{line} \mbox{. It follows from the orthogonality of that } \textcolor{castescuro}{\mbox{circle}} \mbox{ to the two}$
$\small  \textcolor{Blue}{\mbox{initial circles}} \mbox{ that } \textbf\textcolor{castclaro}{\mbox{brown line}} \textbf\mbox{ is orthogonal to the }\textbf\textcolor{verde}{\mbox{inverses (green)}}.$
$\small \mbox{But line  } \textbf\mbox{AB is also perpendicular to } \textbf\textcolor{verde}{\mbox{green circles}} \mbox{ since it is }$
$\small \mbox{orthogonal to the } \textcolor{Blue}{\mbox{initial circles }} \mbox{and the inverse of AB coincides with itself.}$ 

$\small \mbox{If O is one of the intersection points of the } \textcolor{castescuro}{\mbox{new circle}} \mbox{ with line AB, we are}$
$\small \mbox{going to use a second inversion in the plane, defined by } \textcolor{rosa}{circle \;C_{2} } \mbox{ with center}$
$\small \mbox{O (whose radius you may manipulate with the new cursor).}$

$\small \definecolor{verde}{rgb}{0,0.8,0} \textcolor{verde}{\mbox{The inverses}} \mbox{ of the } \textcolor{blue}{\mbox{initial circles}} \mbox{ with respect to }\textcolor{rosa}{circle\;C_{2}} \mbox{ are drawn in } $
$\small\definecolor{verde}{rgb}{0,0.8,0} \textcolor{verde}{\mbox{green}}. $

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$\small\mbox{The } \definecolor{castclaro}{rgb}{0.8,0.6,0} \textbf\textcolor{castclaro}{circle} \mbox{ with center M that passes through L }\mbox{\textbf{is orthogonal to the two}}$ 
$\small\textbf\textcolor{ciano}{\mbox{light blue objects}} \mbox{: in the case of the } \textcolor{ciano}{line} \mbox{ because it passes through the}$
$\small\mbox{center of the }  \small\textcolor{castclaro}{circle} \small\mbox{; in the case of the } \textcolor{ciano}{circle} \mbox{ because LG is perpendicular to } $
$\small\mbox{LM. Moreover, the } \small\textbf\textcolor{castclaro}{\mbox{brown circle }}  \textbf\mbox{intersects r in two points} \mbox{: its radius}$
$\small\mbox{(}\overline{LM}) \mbox{ is bigger than the distance of M to r, since the  }\textcolor{ciano}{\mbox{light blue circle}} \mbox{ is}$ 
$\small\mbox{disjoint from the  } \textcolor{ciano}{\mbox{light blue line}}.$

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$\small \mbox{We saw in the preceding step that there are two two lines orthogonal to}$ 
$\small \mbox{both } \textcolor{verde}{\mbox{green circles}} \mbox{. But a straight line is orthogonal to a circle if and}$
$\small \mbox{only if it passes through the center. Therefore, both lines pass through}$
$\small \mbox{the centers of the } \textcolor{verde}{\mbox{green circles}} \mbox{, which is only possible if the centers do}$
$\small \mbox{coincide. Conclusion: the}\textcolor{verde}{\mbox{ green circles}} \mbox{ are concentric. }$

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A = (4.378753283800812, -14.10690382518224)
B = (8.944055971773547, -15.593333455710665)
r_{1} = 8.8 (slider from 0 to 10, step 0.1)
r_{2} = 2.6 (slider from 0 to 10, step 0.1)
c = Circle(A, r_{1})
d = Circle(B, r_{2})
r = Line(A, B)
E = Intersect(c, r, 1)
F = Intersect(d, r, 1)
C_{1} = Circle(E, F)
d' = Mirror(d, C_{1})
t = Mirror(c, C_{1})
C = Point(d')
D = Point(d')
b = Segment(C, D)
f = LineBisector(C, D)
G = Intersect(r, f)
H = Point(t)
I = Point(t)
h = Line(H, I)
i_1 = Line(G, h)
L = Intersect(d', i_1, 1)
j = Line(L, r)
M = Intersect(t, j, 1)
k = Circle(M, L)
k' = Mirror(k, C_{1})
r_{3} = 4.9 (slider from 0 to 10, step 0.1)
O = Intersect(k', r, 1)
C_{2} = Circle(O, r_{3})
c'_1 = Mirror(c, C_{2})
d'_1 = Mirror(d, C_{2})
i = 0 (slider from 0 to 13, step 1)
distânciaAB = Distance(A, B)
TextoAB: Name[A] + (Name[B]) + " = " + distânciaAB
distânciaAE = Distance(A, E)
a = false
e = false
k'' = Mirror(k', C_{2})
J, K = Intersect(k'', r)