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Question Number 53252 by ajfour last updated on 19/Jan/19
Commented by ajfour last updated on 19/Jan/19
Find a and r in terms of b.
$${Find}\:{a}\:{and}\:{r}\:{in}\:{terms}\:{of}\:{b}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19
centreA is (x−0)^2 +(y−b)^2 =a^2 centre B is (x−a−b)^2 +(y−b)^2 =b^2 AC=(√((r−0)^2 +(r−b)^2 )) =r+a r^2 +r^2 −2rb+b^2 =r^2 +2ra+a^2 r^2 −2rb+b^2 −2ra−a^2 =0....eqn 1 BC (a+b−r)^2 +(b−r)^2 =(r+b)^2 a^2 +b^2 +r^2 +2ab−2ar−2br+b^2 −2br+r^2 =r^2 +2br+b^2 a^2 +2b^2 +2r^2 +2ab−2ar−4br−r^2 −2br−b^2 =0 a^2 +b^2 +r^2 +2ab−2ar−6br=0....eqn 2 r^2 −2ra−a^2 =2rb−b^2 frlm 1 r^2 −2ar+a^2 =6br−2ab−b^2 from 2 r^2 −2ra+b^2 =a^2 +2rb r^2 −2ra+b^2 =6br−2ab−a^2 so a^2 +2rb=6br−2ab−a^2 2a^2 −4br+2ab=0 r=((2a^2 +2ab)/(4b))=((a(a+b))/(2b)) ← r^2 −2ar−a^2 =2rb−b^2 (((a^2 +ab)/(2b)))^2 −2(a+b)(((a^2 +ab)/(2b)))=a^2 −b^2 ((a^2 (a+b)^2 )/(4b^2 ))−(((a+b)(a+b)a)/b)=(a+b)(a−b) deviding by a+b both side ((a^2 (a+b))/(4b^2 ))−((a(a+b))/b)=a−b (a+b)((a^2 /(4b^2 ))−(a/b))=a−b (a^2 /(4b^2 ))−(a/b)=((a−b)/(a+b)) let (a/b)=k (k^2 /4)−k=((k−1)/(k+1)) ((k^2 −4k)/4)=((k−1)/(k+1)) k^3 −4k^2 +k^2 −4k=4k−4 k^3 −3k^2 −8k+4=0
$$ \\ $$$${centreA}\:{is}\:\left({x}−\mathrm{0}\right)^{\mathrm{2}} +\left({y}−{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${centre}\:{B}\:{is}\:\left({x}−{a}−{b}\right)^{\mathrm{2}} +\left({y}−{b}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$${AC}=\sqrt{\left({r}−\mathrm{0}\right)^{\mathrm{2}} +\left({r}−{b}\right)^{\mathrm{2}} }\:={r}+{a} \\ $$$${r}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{rb}+{b}^{\mathrm{2}} ={r}^{\mathrm{2}} +\mathrm{2}{ra}+{a}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{rb}+{b}^{\mathrm{2}} −\mathrm{2}{ra}−{a}^{\mathrm{2}} =\mathrm{0}….{eqn}\:\mathrm{1} \\ $$$${BC} \\ $$$$\left({a}+{b}−{r}\right)^{\mathrm{2}} +\left({b}−{r}\right)^{\mathrm{2}} =\left({r}+{b}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{ab}−\mathrm{2}{ar}−\mathrm{2}{br}+{b}^{\mathrm{2}} −\mathrm{2}{br}+{r}^{\mathrm{2}} ={r}^{\mathrm{2}} +\mathrm{2}{br}+{b}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} +\mathrm{2}{ab}−\mathrm{2}{ar}−\mathrm{4}{br}−{r}^{\mathrm{2}} −\mathrm{2}{br}−{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{ab}−\mathrm{2}{ar}−\mathrm{6}{br}=\mathrm{0}….{eqn}\:\mathrm{2} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{ra}−{a}^{\mathrm{2}} =\mathrm{2}{rb}−{b}^{\mathrm{2}} \:{frlm}\:\mathrm{1} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{ar}+{a}^{\mathrm{2}} =\mathrm{6}{br}−\mathrm{2}{ab}−{b}^{\mathrm{2}} \:\:{from}\:\mathrm{2} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{ra}+{b}^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{2}{rb} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{ra}+{b}^{\mathrm{2}} =\mathrm{6}{br}−\mathrm{2}{ab}−{a}^{\mathrm{2}} \\ $$$${so}\:\:{a}^{\mathrm{2}} +\mathrm{2}{rb}=\mathrm{6}{br}−\mathrm{2}{ab}−{a}^{\mathrm{2}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} −\mathrm{4}{br}+\mathrm{2}{ab}=\mathrm{0} \\ $$$${r}=\frac{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{ab}}{\mathrm{4}{b}}=\frac{{a}\left({a}+{b}\right)}{\mathrm{2}{b}}\:\leftarrow \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{ar}−{a}^{\mathrm{2}} =\mathrm{2}{rb}−{b}^{\mathrm{2}} \\ $$$$\left(\frac{{a}^{\mathrm{2}} +{ab}}{\mathrm{2}{b}}\right)^{\mathrm{2}} −\mathrm{2}\left({a}+{b}\right)\left(\frac{{a}^{\mathrm{2}} +{ab}}{\mathrm{2}{b}}\right)={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$$\frac{{a}^{\mathrm{2}} \left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} }−\frac{\left({a}+{b}\right)\left({a}+{b}\right){a}}{{b}}=\left({a}+{b}\right)\left({a}−{b}\right) \\ $$$${deviding}\:{by}\:{a}+{b}\:{both}\:{side} \\ $$$$\frac{{a}^{\mathrm{2}} \left({a}+{b}\right)}{\mathrm{4}{b}^{\mathrm{2}} }−\frac{{a}\left({a}+{b}\right)}{{b}}={a}−{b} \\ $$$$\left({a}+{b}\right)\left(\frac{{a}^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} }−\frac{{a}}{{b}}\right)={a}−{b} \\ $$$$\frac{{a}^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} }−\frac{{a}}{{b}}=\frac{{a}−{b}}{{a}+{b}}\: \\ $$$${let}\:\frac{{a}}{{b}}={k} \\ $$$$\frac{{k}^{\mathrm{2}} }{\mathrm{4}}−{k}=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$$$\frac{{k}^{\mathrm{2}} −\mathrm{4}{k}}{\mathrm{4}}=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$$${k}^{\mathrm{3}} −\mathrm{4}{k}^{\mathrm{2}} +{k}^{\mathrm{2}} −\mathrm{4}{k}=\mathrm{4}{k}−\mathrm{4} \\ $$$${k}^{\mathrm{3}} −\mathrm{3}{k}^{\mathrm{2}} −\mathrm{8}{k}+\mathrm{4}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19
Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19
k=(a/b)=0.438 so a=0.438b (a/b)=4.562 a=4.562b →this solution not feasible from diagram...
$${k}=\frac{{a}}{{b}}=\mathrm{0}.\mathrm{438}\:\:\:{so}\:\:\:{a}=\mathrm{0}.\mathrm{438}{b} \\ $$$$\frac{{a}}{{b}}=\mathrm{4}.\mathrm{562}\:\:\:{a}=\mathrm{4}.\mathrm{562}{b}\:\rightarrow{this}\:{solution}\:{not}\:{feasible} \\ $$$${from}\:{diagram}… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19
r=((a(a+b))/(2b))=((0.438b(0.438b+b))/(2b))=0.219(1.438b) r=0.3149b
$${r}=\frac{{a}\left({a}+{b}\right)}{\mathrm{2}{b}}=\frac{\mathrm{0}.\mathrm{438}{b}\left(\mathrm{0}.\mathrm{438}{b}+{b}\right)}{\mathrm{2}{b}}=\mathrm{0}.\mathrm{219}\left(\mathrm{1}.\mathrm{438}{b}\right) \\ $$$${r}=\mathrm{0}.\mathrm{3149}{b} \\ $$
Answered by ajfour last updated on 19/Jan/19
i obtain a= b((√2)−1) r= (2+(√2)−2(√(1+(√2))) )b .
$${i}\:{obtain} \\ $$$$\:\:\:\:\:{a}=\:{b}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:{r}=\:\left(\mathrm{2}+\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}\:\right){b}\:. \\ $$
Answered by ajfour last updated on 19/Jan/19
From △AEC_(−) (b−r)^2 +r^2 = (a+r)^2 ...(i) BC^2 =(b+r)^2 = (a+b−r)^2 +(b−r)^2 ....(ii) ⇒ r^2 −2(a+b)r+b^2 −a^2 = 0 & r^2 −2(a+3b)r+(a+b)^2 = 0 comparing the two ((a+b)/(a+3b)) = ((b−a)/(a+b)) ⇒ a^2 +2ab+b^2 = 3b^2 −a^2 −2ab or a^2 +2ab+b^2 = 2b^2 ⇒ a+b = b(√2) a = ((√2)−1)b . now picking up the 2^(nd) quadratic in r r^2 −2(a+3b)r+(a+b)^2 = 0 but we found a=(√2)b−b , so r^2 −2(b(√2)−b+3b)r+2b^2 = 0 r^2 −2(2+(√2))b+2b^2 = 0 ⇒ r =( 2+(√2)±(√((2+(√2))^2 −2)) )b feasible one for diagram is r = (2+(√2)−2(√(1+(√2))) )b .
$$\underset{−} {{From}\:\bigtriangleup{AEC}} \\ $$$$\left({b}−{r}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} =\:\left({a}+{r}\right)^{\mathrm{2}} \:\:\:\:…\left({i}\right) \\ $$$$\:{BC}\:^{\mathrm{2}} \:=\left({b}+{r}\right)^{\mathrm{2}} =\:\left({a}+{b}−{r}\right)^{\mathrm{2}} +\left({b}−{r}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({ii}\right) \\ $$$$\Rightarrow\:{r}^{\mathrm{2}} −\mathrm{2}\left({a}+{b}\right){r}+{b}^{\mathrm{2}} −{a}^{\mathrm{2}} =\:\mathrm{0}\:\:\:\& \\ $$$$\:\:\:\:\:\:{r}^{\mathrm{2}} −\mathrm{2}\left({a}+\mathrm{3}{b}\right){r}+\left({a}+{b}\right)^{\mathrm{2}} =\:\mathrm{0} \\ $$$${comparing}\:{the}\:{two} \\ $$$$\:\:\:\:\:\:\frac{{a}+{b}}{{a}+\mathrm{3}{b}}\:=\:\frac{{b}−{a}}{{a}+{b}} \\ $$$$\Rightarrow\:\:\:{a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} =\:\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} −\mathrm{2}{ab} \\ $$$${or}\:\:\:\:{a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} \:=\:\mathrm{2}{b}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:{a}+{b}\:=\:{b}\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:{a}\:=\:\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){b}\:. \\ $$$${now}\:{picking}\:{up}\:{the}\:\mathrm{2}^{{nd}} \:{quadratic}\:{in}\:{r} \\ $$$$\:\:\:\:\:\:{r}^{\mathrm{2}} −\mathrm{2}\left({a}+\mathrm{3}{b}\right){r}+\left({a}+{b}\right)^{\mathrm{2}} =\:\mathrm{0} \\ $$$${but}\:{we}\:{found}\:\:{a}=\sqrt{\mathrm{2}}{b}−{b}\:,\:{so} \\ $$$$\:\:{r}^{\mathrm{2}} −\mathrm{2}\left({b}\sqrt{\mathrm{2}}−{b}+\mathrm{3}{b}\right){r}+\mathrm{2}{b}^{\mathrm{2}} =\:\mathrm{0} \\ $$$$\:\:\:{r}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){b}+\mathrm{2}{b}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:{r}\:=\left(\:\mathrm{2}+\sqrt{\mathrm{2}}\pm\sqrt{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}}\:\right){b} \\ $$$${feasible}\:{one}\:{for}\:{diagram}\:{is}\: \\ $$$$\:\:\:\:\:\:{r}\:=\:\left(\mathrm{2}+\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}\:\right){b}\:\:. \\ $$
Commented by mr W last updated on 20/Jan/19
sir, i don′t think we can get ((a+b)/(a+3b)) = ((b−a)/(a+b)) from r^2 −2(a+b)r+b^2 −a^2 = 0 & r^2 −2(a+3b)r+(a+b)^2 = 0. let′s look at a_1 x^2 +b_1 x+c_1 =0 a_2 x^2 +b_2 x+c_2 =0 if both eqn. have two equal roots, then (a_1 /a_2 )=(b_1 /b_2 )=(c_1 /c_2 ) if both eqn. have only one equal root, we can not say (a_1 /a_2 )=(b_1 /b_2 )=(c_1 /c_2 ) i think we have here the second case.
$${sir},\:{i}\:{don}'{t}\:{think}\:{we}\:{can}\:{get} \\ $$$$\:\:\:\:\:\:\frac{{a}+{b}}{{a}+\mathrm{3}{b}}\:=\:\frac{{b}−{a}}{{a}+{b}} \\ $$$${from} \\ $$$$\:\:\:\:\:\:{r}^{\mathrm{2}} −\mathrm{2}\left({a}+{b}\right){r}+{b}^{\mathrm{2}} −{a}^{\mathrm{2}} =\:\mathrm{0}\:\:\:\& \\ $$$$\:\:\:\:\:\:{r}^{\mathrm{2}} −\mathrm{2}\left({a}+\mathrm{3}{b}\right){r}+\left({a}+{b}\right)^{\mathrm{2}} =\:\mathrm{0}. \\ $$$$ \\ $$$${let}'{s}\:{look}\:{at} \\ $$$${a}_{\mathrm{1}} {x}^{\mathrm{2}} +{b}_{\mathrm{1}} {x}+{c}_{\mathrm{1}} =\mathrm{0} \\ $$$${a}_{\mathrm{2}} {x}^{\mathrm{2}} +{b}_{\mathrm{2}} {x}+{c}_{\mathrm{2}} =\mathrm{0} \\ $$$${if}\:{both}\:{eqn}.\:{have}\:{two}\:{equal}\:{roots}, \\ $$$${then} \\ $$$$\frac{{a}_{\mathrm{1}} }{{a}_{\mathrm{2}} }=\frac{{b}_{\mathrm{1}} }{{b}_{\mathrm{2}} }=\frac{{c}_{\mathrm{1}} }{{c}_{\mathrm{2}} } \\ $$$${if}\:{both}\:{eqn}.\:{have}\:{only}\:{one}\:{equal}\:{root}, \\ $$$${we}\:{can}\:{not}\:{say} \\ $$$$\frac{{a}_{\mathrm{1}} }{{a}_{\mathrm{2}} }=\frac{{b}_{\mathrm{1}} }{{b}_{\mathrm{2}} }=\frac{{c}_{\mathrm{1}} }{{c}_{\mathrm{2}} } \\ $$$${i}\:{think}\:{we}\:{have}\:{here}\:{the}\:{second}\:{case}. \\ $$

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