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

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

$$maybeimpossibletosolve:$$$$findtheradiusofthe{n}^{th}circle.$$

Commented by aleks041103 last updated on 05/Jan/22

$$I^{\prime }lltry...$$

Commented by behi834171 last updated on 07/Jan/22

$${\mathit{r}}_{\mathit{n}}=\frac{1}{2^{(\mathit{n}-1)}}{(2-\sqrt{2})}^{\mathit{n}}{(2-\sqrt{2-\sqrt{2}})}^{(\mathit{n}-1)}.$$$$\text{You can't use 'macro parameter character \#' in math mode}$$

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

$$thankssir!$$$$yes.irememberthisquestionwas$$$$oncediscussed.$$$$buttheformulayoumentionedis$$$$notcorrect.only{r}_{1}iscorrect.upon$$$$n=2,itisnotcorrect,becausethe$$$$circles{don}^{\prime }ttangentwiththecurve$$$$y=\frac{1}{x},seediagram.$$$$ithink{it}^{\prime }simpossibletofinda$$$$formulafortheradiusof{n}^{th}circle$$$$intermsofn.$$

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

$$y^2=x^2+2$$$$y=\sqrt{x^2+2}$$$$\frac{dy}{dx}=\frac{x}{\sqrt{x^2+2}}=$$$$a\sqrt{2}+a=\sqrt{2}\Rightarrow a=2-\sqrt{2}$$$$A[0,2\sqrt{2}-2]$$$$letB[(a+b)\sin \theta ,a\sqrt{2}+(a+b)\cos \theta ]$$$$b=\frac{a\sqrt{2}+(a+b)(\cos \theta -\sin \theta )}{\sqrt{2}}$$$$letCircleBtouchesy=\sqrt{x^2+2}at$$$$T_2[p,\sqrt{p^2+2}]$$$$a\sqrt{2}+(a+b)\cos \theta +b\sin \varphi =\sqrt{p^2+2}$$$$(a+b)\sin \theta -b\cos \varphi =p$$$$subtracting$$$$\Rightarrow a\sqrt{2}+\sqrt{2}(b-a)+b(\sin \varphi +\cos \varphi )$$$$=\sqrt{p^2+2}-p.....\left(i\right)$$$$\tan \varphi =\frac{\sin \varphi }{\cos \varphi }=\frac{p}{\sqrt{p^2+2}}$$$$\Rightarrow p^2=\frac{2}{\cot {}^2\varphi -1}...\left(ii\right)$$$$and{(p+b\cos \varphi )}^2+{(\sqrt{p^2+2}-a\sqrt{2}-b\sin \varphi )}^2$$$$={(a+b)}^2$$$$$$$$\Rightarrow p^2+b^2+2bp\cos \varphi$$$$+{(\sqrt{p^2+2}-a\sqrt{2})}^2$$$$-2b(\sqrt{p^2+2}-a\sqrt{2})\sin \varphi ={(a+b)}^2$$$$......\left(iii\right)$$$$threeeqns.unknowns:b,p,\tan \varphi$$

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