Question-163280

Question Number 163280 by mr W last updated on 05/Jan/22

Commented by mr W last updated on 05/Jan/22

m a y b e i m p o s s i b l e t o s o l v e :

f i n d t h e r a d i u s o f t h e n^{t h} c i r c l e .

Commented by aleks041103 last updated on 05/Jan/22

I^{'} l l t r y \dots

Commented by behi834171 last updated on 07/Jan/22

r_{n} = \frac{1}{2^{(n - 1)}} {(2 - \sqrt{2})}^{n} {(2 - \sqrt{2 - \sqrt{2}})}^{(n - 1)} .

You can't use 'macro parameter character #' in math mode

Commented by mr W last updated on 07/Jan/22

t h a n k s s i r!

y e s . i r e m e m b e r t h i s q u e s t i o n w a s

o n c e d i s c u s s e d .

b u t t h e f o r m u l a y o u m e n t i o n e d i s

n o t c o r r e c t . o n l y r_{1} i s c o r r e c t . u p o n

n = 2, i t i s n o t c o r r e c t, b e c a u s e t h e

c i r c l e s {d o n}^{'} t t a n g e n t w i t h t h e c u r v e

y = \frac{1}{x}, s e e d i a g r a m .

i t h i n k {i t}^{'} s i m p o s s i b l e t o f i n d a

f o r m u l a f o r t h e r a d i u s o f n^{t h} c i r c l e

i n t e r m s o f n .

Commented by mr W last updated on 08/Jan/22

Answered by ajfour last updated on 06/Jan/22

Commented by ajfour last updated on 08/Jan/22

y^{2} = x^{2} + 2

y = \sqrt{x^{2} + 2}

\frac{d y}{d x} = \frac{x}{\sqrt{x^{2} + 2}} =

a \sqrt{2} + a = \sqrt{2} \Rightarrow a = 2 - \sqrt{2}

A [0, 2 \sqrt{2} - 2]

l e t B [(a + b) \sin θ, a \sqrt{2} + (a + b) \cos θ]

b = \frac{a \sqrt{2} + (a + b) (\cos θ - \sin θ)}{\sqrt{2}}

l e t C i r c l e B t o u c h e s y = \sqrt{x^{2} + 2} a t

T_{2} [p, \sqrt{p^{2} + 2}]

a \sqrt{2} + (a + b) \cos θ + b \sin ϕ = \sqrt{p^{2} + 2}

(a + b) \sin θ - b \cos ϕ = p

s u b t r a c t i n g

\Rightarrow a \sqrt{2} + \sqrt{2} (b - a) + b (\sin ϕ + \cos ϕ)

= \sqrt{p^{2} + 2} - p \dots . . (i)

\tan ϕ = \frac{\sin ϕ}{\cos ϕ} = \frac{p}{\sqrt{p^{2} + 2}}

\Rightarrow p^{2} = \frac{2}{\cot^{2} ϕ - 1} \dots (i i)

a n d {(p + b \cos ϕ)}^{2} + {(\sqrt{p^{2} + 2} - a \sqrt{2} - b \sin ϕ)}^{2}

= {(a + b)}^{2}

\Rightarrow p^{2} + b^{2} + 2 b p \cos ϕ

+ {(\sqrt{p^{2} + 2} - a \sqrt{2})}^{2}

- 2 b (\sqrt{p^{2} + 2} - a \sqrt{2}) \sin ϕ = {(a + b)}^{2}

\dots \dots (i i i)

t h r e e e q n s . u n k n o w n s : b, p, \tan ϕ

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