\; \text{Apolloniova kružnica je GMB, pre ktoré platí: } \frac{|P X|}{|P X|}= λ;  \; p=PQ \text{ - pevne daná priamka

 \text{Postup} 

1.  \; R,S∈p:\frac{|P R|}{|QR|}= λ=\frac{|P T|}{|QT|}\text{ - pomocou podobnosti }\triangle 

λ

2.  \; ω=(S_{RS},r=SR)\text{ - \red{Apolloniova kružnica }

\text{ Nech bod } X∈ω \text{ - ľub.}

4.  \; \triangle PRX ∼ \triangle QRF⇒\frac{|PR|}{|QR|}= \frac{|P X|}{|QF|}=λ

5.  \; \triangle PTX ∼ \triangle QTE⇒\frac{|PT|}{|QT|}= \frac{|P X|}{|EQ|}=λ

6. \; \text{Odkiaľ : }\left [ QF \right ] =\left [QE  \right ] ⇒Q \text{ je stred prepony }\triangle EFX

7. \; \text{Kružnica Tálesova v } \triangle EFX: \; \left|QF \right| =\left| QX \right| 

\text{ \pmb{Záver}: dosadením do 4. a po úprave:  }\red{ \frac{ | PX| }{| QX| }=λ}

\text{Dôkaz: }\left [ \frac{ | PX| }{| QX| }=λ⇒X∈ω \right ] \text{ ... pozrite si prácu}

\text{https://dml.cz/bitstream/handle/10338.dmlcz/146126/Rozhledy_081-2006-1_2.pdf

C = (-5.846942700599799, -2.4559221777710154)
P = (-0.6134036859115144, 2.047355579053787)
i = Line(P, C)
λ = 2.5 (slider from 0.1 to 5, step 0.1)
c = Circle(P, λ)
B = Intersect(c, i, 2)
Q = (7.2247338151565215, 2.047355579053787)
f = Line(P, Q)
j = Line(Q, i)
d = Circle(Q, 1)
A: B
g = Segment(P, A)
G = Intersect(c, i, 1)
D = Intersect(d, j, 1)
k = Segment(Q, D)
E_{1} = Intersect(d, j, 2)
l = Segment(Q, E_{1})
m = Line(A, E_{1})
n = Line(A, D)
T = Intersect(f, m)
R = Intersect(f, n)
t1, a, p, r = Polygon(P, T, A)
t2, q, h_1, i_1 = Polygon(T, Q, E_{1})
S = Midpoint(T, R)
ω = Circle(S, T)
X = Point(ω)
h = Line(X, P)
s = Line(Q, h)
t3, t, e, g_1 = Polygon(P, X, R)
b = Segment(X, R)
F = Intersect(s, b)
t4, f_1, q_1, t_1 = Polygon(Q, R, F)
j_1 = Line(X, T)
E = Intersect(s, j_1)
k_1 = Segment(E, R)
t5, x_1, p_1, t_2 = Polygon(P, T, X)
t6, l_1, m_1, n_1 = Polygon(E, T, Q)