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Question-51422

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Question Number 51422 by ajfour last updated on 26/Dec/18
Commented by ajfour last updated on 26/Dec/18
Find R the radius of semicircle in terms of radii a and b of the circles.
$${Find}\:{R}\:{the}\:{radius}\:{of}\:{semicircle} \\ $$$${in}\:{terms}\:{of}\:{radii}\:{a}\:{and}\:{b}\:{of}\:{the} \\ $$$${circles}. \\ $$
Answered by ajfour last updated on 26/Dec/18
let sin θ = (a/(R−a)) , sin φ = (b/(R−b)) (R−a)cos θ+(R−b)cos φ = (√(4ab)) ⇒ (√(R(R−2a)))+(√(R(R−2b))) = (√(4ab)) squaring 2R^2 −2(a+b)R +2(√(R^2 {R^2 −2(a+b)R+4ab})) = 4ab rearranging and squaring R^2 {R^2 −2(a+b)R+4ab} = {2ab−R^2 +(a+b)R}^2 ⇒ 4abR^2 = 4a^2 b^2 +(a+b)^2 R^2 −4abR^2 +4ab(a+b)R ⇒ (a^2 +b^2 −6ab)R^2 +4ab(a+b)R +4a^2 b^2 = 0 ⇒ R = ((−2ab(a+b)−(√(4a^2 b^2 (8ab))))/(a^2 +b^2 −6ab)) or R = [(√(8ab))+(a+b)](((2ab)/(6ab−a^2 −b^2 ))) .
$${let}\:\:\mathrm{sin}\:\theta\:=\:\frac{{a}}{{R}−{a}}\:,\:\mathrm{sin}\:\phi\:=\:\frac{{b}}{{R}−{b}} \\ $$$$\:\:\:\left({R}−{a}\right)\mathrm{cos}\:\theta+\left({R}−{b}\right)\mathrm{cos}\:\phi\:=\:\sqrt{\mathrm{4}{ab}} \\ $$$$\Rightarrow\:\sqrt{{R}\left({R}−\mathrm{2}{a}\right)}+\sqrt{{R}\left({R}−\mathrm{2}{b}\right)}\:=\:\sqrt{\mathrm{4}{ab}} \\ $$$$\:{squaring} \\ $$$$\:\:\mathrm{2}{R}^{\mathrm{2}} −\mathrm{2}\left({a}+{b}\right){R}\:+\mathrm{2}\sqrt{{R}^{\mathrm{2}} \left\{{R}^{\mathrm{2}} −\mathrm{2}\left({a}+{b}\right){R}+\mathrm{4}{ab}\right\}}\:=\:\mathrm{4}{ab} \\ $$$${rearranging}\:{and}\:{squaring} \\ $$$${R}^{\mathrm{2}} \left\{{R}^{\mathrm{2}} −\mathrm{2}\left({a}+{b}\right){R}+\mathrm{4}{ab}\right\} \\ $$$$\:\:\:\:\:=\:\left\{\mathrm{2}{ab}−{R}^{\mathrm{2}} +\left({a}+{b}\right){R}\right\}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{4}{abR}^{\mathrm{2}} \:=\:\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\left({a}+{b}\right)^{\mathrm{2}} {R}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:−\mathrm{4}{abR}^{\mathrm{2}} +\mathrm{4}{ab}\left({a}+{b}\right){R} \\ $$$$\Rightarrow\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{6}{ab}\right){R}^{\mathrm{2}} +\mathrm{4}{ab}\left({a}+{b}\right){R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:\boldsymbol{{R}}\:=\:\frac{−\mathrm{2}\boldsymbol{{ab}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)−\sqrt{\mathrm{4}\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{b}}^{\mathrm{2}} \left(\mathrm{8}\boldsymbol{{ab}}\right)}}{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} −\mathrm{6}\boldsymbol{{ab}}} \\ $$$${or}\:\:{R}\:=\:\left[\sqrt{\mathrm{8}{ab}}+\left({a}+{b}\right)\right]\left(\frac{\mathrm{2}{ab}}{\mathrm{6}{ab}−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)\:. \\ $$
Answered by behi83417@gmail.com last updated on 27/Dec/18
⇒AB=(√(2(a^2 +b^2 ))) TO=OS=R OA=R−a,OB=R−b sinα=(a/(R−a)),sinβ=(b/(R−b)) cosα=((√((R−a)^2 −a^2 ))/(R−a)),cosβ=((√((R−b)^2 −b^2 ))/(R−b)) cosϕ=cos(π−α−β)=−cos(α+β)= =((ab−(√(R^2 −2Ra)).(√(R^2 −2Rb)))/((R−a)(R−b)))(i) cosϕ=((OA^2 +OB^2 −AB^2 )/(2OA.OB))=(((R−a)^2 +(R−b)^2 −2(a^2 +b^2 ))/(2(R−a)(R−b)))= =((2R^2 −2Ra−2Rb−a^2 −b^2 )/(2(R−a)(R−b)))(ii) i=ii⇒ 2R^2 −2Ra−2Rb−a^2 −b^2 = =ab−(√(R^2 −2Ra)).(√(R^2 −2Rb)) a^2 +b^2 +ab+2Ra+2Rb−2R^2 =(√(R^2 −2Ra)).(√(R^2 −2Rb)) R^2 −2Ra=p,R^2 −2Rb=q,a^2 +b^2 +ab=t a^2 +b^2 +ab−(p+q)=(√(pq)) t−(p+q)=(√(pq))⇒t^2 −2t(p+q)+(p+q)^2 =pq p+q=2R^2 −2Ra−2Rb (p+q)^2 =4R^4 +4R^2 a^2 +4R^2 b^2 −8R^3 a−8R^3 b+8R^2 ab pq=(R^2 −2Ra)(R^2 −2Rb)=R^4 −2R^3 b−2R^3 a+4R^2 ab t^2 −4R(R−a−b)t+3R^4 +4R^2 (a^2 +b^2 +ab) −6R^3 a−6R^3 b=0 t^2 +4R(a+b)t+3R^3 (R−2a−2b)=0 ⇒△′=4R^2 (a+b)^2 −3R^3 (R−2a−2b)= =R^2 [4(a+b)^2 −3R^2 +6R(a+b)]=R^2 m^2 a^2 +b^2 +ab=−2R(a+b)+Rm m=((a^2 +b^2 +ab)/R)+2(a+b)
$$ \\ $$$$\Rightarrow{AB}=\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$$${TO}={OS}={R} \\ $$$${OA}={R}−{a},{OB}={R}−{b} \\ $$$${sin}\alpha=\frac{{a}}{{R}−{a}},{sin}\beta=\frac{{b}}{{R}−{b}} \\ $$$${cos}\alpha=\frac{\sqrt{\left({R}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{R}−{a}},{cos}\beta=\frac{\sqrt{\left({R}−{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{R}−{b}} \\ $$$${cos}\varphi={cos}\left(\pi−\alpha−\beta\right)=−{cos}\left(\alpha+\beta\right)= \\ $$$$=\frac{{ab}−\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}.\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}}}{\left({R}−{a}\right)\left({R}−{b}\right)}\left({i}\right) \\ $$$${cos}\varphi=\frac{{OA}^{\mathrm{2}} +{OB}^{\mathrm{2}} −{AB}^{\mathrm{2}} }{\mathrm{2}{OA}.{OB}}=\frac{\left({R}−{a}\right)^{\mathrm{2}} +\left({R}−{b}\right)^{\mathrm{2}} −\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{\mathrm{2}\left({R}−{a}\right)\left({R}−{b}\right)}= \\ $$$$=\frac{\mathrm{2}{R}^{\mathrm{2}} −\mathrm{2}{Ra}−\mathrm{2}{Rb}−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}\left({R}−{a}\right)\left({R}−{b}\right)}\left({ii}\right) \\ $$$${i}={ii}\Rightarrow \\ $$$$\mathrm{2}{R}^{\mathrm{2}} −\mathrm{2}{Ra}−\mathrm{2}{Rb}−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} = \\ $$$$={ab}−\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}.\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}+\mathrm{2}{Ra}+\mathrm{2}{Rb}−\mathrm{2}{R}^{\mathrm{2}} =\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Ra}}.\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}} \\ $$$${R}^{\mathrm{2}} −\mathrm{2}{Ra}={p},{R}^{\mathrm{2}} −\mathrm{2}{Rb}={q},{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}={t} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}−\left({p}+{q}\right)=\sqrt{{pq}} \\ $$$${t}−\left({p}+{q}\right)=\sqrt{{pq}}\Rightarrow{t}^{\mathrm{2}} −\mathrm{2}{t}\left({p}+{q}\right)+\left({p}+{q}\right)^{\mathrm{2}} ={pq} \\ $$$${p}+{q}=\mathrm{2}{R}^{\mathrm{2}} −\mathrm{2}{Ra}−\mathrm{2}{Rb} \\ $$$$\left({p}+{q}\right)^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{4}} +\mathrm{4}{R}^{\mathrm{2}} {a}^{\mathrm{2}} +\mathrm{4}{R}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{8}{R}^{\mathrm{3}} {a}−\mathrm{8}{R}^{\mathrm{3}} {b}+\mathrm{8}{R}^{\mathrm{2}} {ab} \\ $$$${pq}=\left({R}^{\mathrm{2}} −\mathrm{2}{Ra}\right)\left({R}^{\mathrm{2}} −\mathrm{2}{Rb}\right)={R}^{\mathrm{4}} −\mathrm{2}{R}^{\mathrm{3}} {b}−\mathrm{2}{R}^{\mathrm{3}} {a}+\mathrm{4}{R}^{\mathrm{2}} {ab} \\ $$$${t}^{\mathrm{2}} −\mathrm{4}{R}\left({R}−{a}−{b}\right){t}+\mathrm{3}{R}^{\mathrm{4}} +\mathrm{4}{R}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}\right) \\ $$$$−\mathrm{6}{R}^{\mathrm{3}} {a}−\mathrm{6}{R}^{\mathrm{3}} {b}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} +\mathrm{4}{R}\left({a}+{b}\right){t}+\mathrm{3}{R}^{\mathrm{3}} \left({R}−\mathrm{2}{a}−\mathrm{2}{b}\right)=\mathrm{0} \\ $$$$\Rightarrow\bigtriangleup'=\mathrm{4}{R}^{\mathrm{2}} \left({a}+{b}\right)^{\mathrm{2}} −\mathrm{3}{R}^{\mathrm{3}} \left({R}−\mathrm{2}{a}−\mathrm{2}{b}\right)= \\ $$$$={R}^{\mathrm{2}} \left[\mathrm{4}\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{3}{R}^{\mathrm{2}} +\mathrm{6}{R}\left({a}+{b}\right)\right]={R}^{\mathrm{2}} {m}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}=−\mathrm{2}{R}\left({a}+{b}\right)+{Rm} \\ $$$${m}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}}{{R}}+\mathrm{2}\left({a}+{b}\right) \\ $$
Commented by ajfour last updated on 27/Dec/18
for b=a , we must get R=a(1+(√2)) your solution gets m= ((3a^2 )/R)+4a 3a^2 =−4aR+3a^2 +4aR so we dont obtain R, please resolve Sir.
$${for}\:{b}={a}\:\:,\:{we}\:{must}\:{get}\:\boldsymbol{{R}}=\boldsymbol{{a}}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$${your}\:{solution}\:{gets} \\ $$$${m}=\:\frac{\mathrm{3}{a}^{\mathrm{2}} }{{R}}+\mathrm{4}{a} \\ $$$$\mathrm{3}{a}^{\mathrm{2}} =−\mathrm{4}{aR}+\mathrm{3}{a}^{\mathrm{2}} +\mathrm{4}{aR} \\ $$$${so}\:{we}\:{dont}\:{obtain}\:{R},\:{please}\:{resolve} \\ $$$${Sir}. \\ $$

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