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Question Number 49736 by behi83417@gmail.com last updated on 10/Dec/18
there is two small and one grater circles that [two]are tangent to [one]and all three circles are inscribed in an ellipse with: [(a/b)=2(√2)]and tangent to it at two points such that center of circles are on major axe of ellipse. find: ((radi of great circle)/(radi of small circle)) .
$$\boldsymbol{\mathrm{there}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{two}}\:\boldsymbol{\mathrm{small}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{grater}}\:\boldsymbol{\mathrm{circles}} \\ $$$$\boldsymbol{\mathrm{that}}\:\left[\boldsymbol{\mathrm{two}}\right]\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{tangent}}\:\boldsymbol{\mathrm{to}}\:\left[\boldsymbol{\mathrm{one}}\right]\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{three}} \\ $$$$\:\boldsymbol{\mathrm{circles}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{inscribed}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{ellipse}}\:\boldsymbol{\mathrm{with}}: \\ $$$$\left[\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}}=\mathrm{2}\sqrt{\mathrm{2}}\right]\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{tangent}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{two}}\:\boldsymbol{\mathrm{points}}\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{center}}\: \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{circles}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{major}}\:\boldsymbol{\mathrm{axe}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{ellipse}}. \\ $$$$\boldsymbol{\mathrm{find}}:\:\:\:\:\frac{\boldsymbol{\mathrm{radi}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{great}}\:\boldsymbol{\mathrm{circle}}}{\boldsymbol{\mathrm{radi}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{small}}\:\boldsymbol{\mathrm{circle}}}\:\:. \\ $$
Commented by ajfour last updated on 10/Dec/18
Answered by ajfour last updated on 10/Dec/18
let b=1, a= 2(√2) eq. of ellipse : (x^2 /8)+y^2 = 1 eq. of larger circle: x^2 +y^2 = 1 eq. of smaller circle: (x−r−1)^2 +y^2 = r^2 for tangency with ellipse, (x−r−1)^2 +(1−(x^2 /8))= r^2 has only one root. ⇒ (7/8)x^2 −2(r+1)x+2(r+1) = 0 or 7x^2 −16(r+1)x+16(r+1)= 0 ⇒ 16×16(r+1)^2 = 7×4×16(r+1) ⇒ 4r+4 = 7 ⇒ r = (3/4) ((Radius of greater circle)/(Radius of smaller circle)) = (1/(3/4)) =(4/3) .
$${let}\:{b}=\mathrm{1},\:\:{a}=\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${eq}.\:{of}\:{ellipse}\::\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{8}}+{y}^{\mathrm{2}} =\:\mathrm{1} \\ $$$${eq}.\:{of}\:{larger}\:{circle}:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\:\mathrm{1} \\ $$$${eq}.\:{of}\:{smaller}\:{circle}: \\ $$$$\:\:\:\:\:\:\:\:\left({x}−{r}−\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\:{r}^{\mathrm{2}} \\ $$$${for}\:{tangency}\:{with}\:{ellipse}, \\ $$$$\:\:\:\:\left({x}−{r}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{8}}\right)=\:{r}^{\mathrm{2}} \:\:{has} \\ $$$${only}\:{one}\:{root}.\:\Rightarrow \\ $$$$\:\:\:\:\:\:\frac{\mathrm{7}}{\mathrm{8}}{x}^{\mathrm{2}} −\mathrm{2}\left({r}+\mathrm{1}\right){x}+\mathrm{2}\left({r}+\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$${or}\:\:\mathrm{7}{x}^{\mathrm{2}} −\mathrm{16}\left({r}+\mathrm{1}\right)\boldsymbol{{x}}+\mathrm{16}\left({r}+\mathrm{1}\right)=\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{16}×\mathrm{16}\left({r}+\mathrm{1}\right)^{\mathrm{2}} =\:\mathrm{7}×\mathrm{4}×\mathrm{16}\left({r}+\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\:\mathrm{4}{r}+\mathrm{4}\:=\:\mathrm{7}\:\:\:\:\Rightarrow\:\:\:{r}\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{{Radius}\:{of}\:{greater}\:{circle}}{{Radius}\:{of}\:{smaller}\:{circle}}\:=\:\frac{\mathrm{1}}{\mathrm{3}/\mathrm{4}}\:=\frac{\mathrm{4}}{\mathrm{3}}\:. \\ $$
Commented by behi83417@gmail.com last updated on 10/Dec/18
thanks a lot sir Ajfour. can you draw a proper picture for this question please?i can′t install lekh diagram. in 4th line from end:x is missing.
$${thanks}\:{a}\:{lot}\:{sir}\:{Ajfour}. \\ $$$${can}\:{you}\:{draw}\:{a}\:{proper}\:{picture}\:{for}\:{this} \\ $$$${question}\:{please}?{i}\:{can}'{t}\:{install}\:\boldsymbol{\mathrm{lekh}}\:\boldsymbol{\mathrm{diagram}}. \\ $$$${in}\:\mathrm{4}{th}\:{line}\:{from}\:{end}:\boldsymbol{{x}}\:{is}\:{missing}. \\ $$
Commented by ajfour last updated on 10/Dec/18
yes Sir, but whats the trouble in installing Lekh Diagram ?
$${yes}\:{Sir},\:{but}\:{whats}\:{the}\:{trouble}\:{in} \\ $$$${installing}\:{Lekh}\:{Diagram}\:? \\ $$
Commented by behi83417@gmail.com last updated on 10/Dec/18
thank you very much sir. it is not available in google play for me.
$${thank}\:{you}\:{very}\:{much}\:{sir}. \\ $$$${it}\:{is}\:{not}\:{available}\:{in}\:{google}\:{play}\:{for}\:{me}. \\ $$

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