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Question Number 10455 by amir last updated on 10/Feb/17

Commented by mrW1 last updated on 10/Feb/17
$circle C1 with radius r_1 and centre point M_1 (r_1 ,h_1 )with h_1 =r_1 has two possibilities: r_1 = { ((2−(√2))),((2+(√2))) :} circle C2 with radius r_2 and centre point M_2 (r_2 ,h_2 ) (h_2 −h_1 )^2 +(r_1 −r_2 )^2 =(r_1 +r_2 )^2 ⇒h_2 =h_1 +2(√(r_1 r_2 ))=r_1 +2(√(r_1 r_2 )) equation of circle C2 is (x−r_2 )^2 +(y−h_2 )^2 =r_2 ^2 (x−r_2 )^2 +(y−r_1 −2(√(r_1 r_2 )))^2 =r_2 ^2 it tangents the curve xy=1 (x−r_2 )^2 +((1/x)−r_1 −2(√(r_1 r_2 )))^2 =r_2 ^2 ...(i) way 1: solve x in terms of r_2 (I know this is not easy) since there is only one solution for x (tangent!) ⇒solution for r_2 ⇒determine h_2 way 2: solve r_2 in terms of x (I know this is not easy) since r_2 must be minimum (tangent!) ⇒(dr_2 /dx)=0 ⇒solve x ⇒determine r_2 ⇒determine h_2 with r_2 and h_2 you can determine r_3 and h_3 for circle C3 in similary way... etc... I don′t think there exists a general formula for circle C_n (n=2,3,...20,...)$
$c i r c l e C 1 w i t h r a d i u s r_{1} a n d c e n t r e$ $p o i n t M_{1} (r_{1}, h_{1}) w i t h h_{1} = r_{1} h a s t w o$ $p o s s i b i l i t i e s :$ $r_{1} = {\begin{cases} 2 - \sqrt{2} \\ 2 + \sqrt{2} \end{cases}$ $c i r c l e C 2 w i t h r a d i u s r_{2} a n d c e n t r e$ $p o i n t M_{2} (r_{2}, h_{2})$ ${(h_{2} - h_{1})}^{2} + {(r_{1} - r_{2})}^{2} = {(r_{1} + r_{2})}^{2}$ $\Rightarrow h_{2} = h_{1} + 2 \sqrt{r_{1} r_{2}} = r_{1} + 2 \sqrt{r_{1} r_{2}}$ $e q u a t i o n o f c i r c l e C 2 i s$ ${(x - r_{2})}^{2} + {(y - h_{2})}^{2} = r_{2}^{2}$ ${(x - r_{2})}^{2} + {(y - r_{1} - 2 \sqrt{r_{1} r_{2}})}^{2} = r_{2}^{2}$ $i t t a n g e n t s t h e c u r v e x y = 1$ ${(x - r_{2})}^{2} + {(\frac{1}{x} - r_{1} - 2 \sqrt{r_{1} r_{2}})}^{2} = r_{2}^{2} . . . (i)$ $w a y 1 :$ $s o l v e x i n t e r m s o f r_{2} (I k n o w t h i s i s n o t e a s y)$ $s i n c e t h e r e i s o n l y o n e s o l u t i o n f o r x (t a n g e n t!)$ $\Rightarrow s o l u t i o n f o r r_{2}$ $\Rightarrow d e t e r m i n e h_{2}$ $w a y 2 :$ $s o l v e r_{2} i n t e r m s o f x (I k n o w t h i s i s n o t e a s y)$ $s i n c e r_{2} m u s t b e m i n i m u m (t a n g e n t!)$ $\Rightarrow \frac{{d r}_{2}}{d x} = 0$ $\Rightarrow s o l v e x$ $\Rightarrow d e t e r m i n e r_{2}$ $\Rightarrow d e t e r m i n e h_{2}$ $w i t h r_{2} a n d h_{2} y o u c a n d e t e r m i n e$ $r_{3} a n d h_{3} f o r c i r c l e C 3 i n s i m i l a r y w a y . . .$ $e t c . . .$ $I {d o n}^{'} t t h i n k t h e r e e x i s t s a g e n e r a l$ $f o r m u l a f o r c i r c l e C_{n} (n = 2, 3, . . . 20, . . .)$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/Feb/17

$h e l l o d e a r m r W 1 . p l e a s e d r a w a$ $d i a g r a m i n l a r g e s c a l e a n d p o s t i t$ $h e r e . i w i l l s o l v e t h i s q u i e s t i o n .$ $t n x 4 a l l .$
Commented by mrW1 last updated on 14/Feb/17

Commented by mrW1 last updated on 14/Feb/17

	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Mar/17

	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Mar/17

	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Mar/17

	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Mar/17

	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Mar/17

	Commented by mrW1 last updated on 16/Mar/17

	$I {c a n}^{'} t s e e h o w i t i s c o n s i d e r e d t h a t$ $t h e c i r c l e s C_{2} a n d s o o n t a n g e n t t h e$ $c i r v e x y = 1 .$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Mar/17

	$i t i s a p p l a y t o f i n d r_{1} .$
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