Question Number 41766 by ajfour last updated on 12/Aug/18
Answered by MrW3 last updated on 18/Aug/18
![let′s say 0<b≤a≤2R B(0,0) O((b/2),(√(R^2 −(b^2 /4)))) C(h,r) Eqn. of BA: y=mx with m=tan (cos^(−1) (b/(2R))±cos^(−1) (a/(2R))) Eqn. of circle with r: (x−h)^2 +(y−r)^2 =r^2 ⇒(x−h)+(y−r)(dy/dx)=0 at M: (x_M −h)+(y_M −r)m=0 (x_M −h)^2 +(y_M −r)^2 =r^2 ⇒(m^2 +1)(x_M −h)^2 =m^2 r^2 ⇒x_M =h±((mr)/( (√(m^2 +1)))) ⇒y_M =r∓(r/( (√(m^2 +1)))) y_M =mx_M ⇒y_M =r∓(r/( (√(m^2 +1))))=mh±((m^2 r)/( (√(m^2 +1)))) ⇒1∓(1/( (√(m^2 +1))))=((mh)/r)±(m^2 /( (√(m^2 +1)))) ⇒h=(r/m)(1±(√(m^2 +1)))=λr>0 with λ=(1/m)(1+(√(m^2 +1))) OC=R−r (√((h−(b/2))^2 +(r−(√(R^2 −(b^2 /4))))^2 ))=R−r (λr−(b/2))^2 +(r−(√(R^2 −(b^2 /4))))^2 =(R−r)^2 λ^2 r^2 +(b^2 /4)−λbr+r^2 +R^2 −(b^2 /4)−2r(√(R^2 −(b^2 /4)))=R^2 +r^2 −2Rr λ^2 r=λb+2(√(R^2 −(b^2 /4)))−2R ⇒r=(1/λ^2 )[λb−2(R−(√(R^2 −(b^2 /4))))] with λ=(1/m)(1+(√(m^2 +1))) and m=tan (cos^(−1) (b/(2R))±cos^(−1) (a/(2R))) =(((√(1−(b^2 /(4R^2 ))))×(a/(2R))±(b/(2R))×(√(1−(a^2 /(4R^2 )))))/((b/(2R))×(a/(2R))∓(√(1−(b^2 /(4R^2 ))))×(√(1−(a^2 /(4R^2 )))))) =((a(√(4R^2 −b^2 ))±b(√(4R^2 −a^2 )))/(ab∓(√((4R^2 −a^2 )(4R^2 −b^2 ))))) with α=(a/(2R))≤1, β=(b/(2R))≤1 and β≤α, m=((α(√(1−β^2 ))±β(√(1−α^2 )))/(αβ∓(√((1−α^2 )(1−β^2 ))))) λ=(1/m)(1+(√(m^2 +1))) ⇒r=((2R)/λ^2 )(λβ+(√(1−β^2 ))−1)](https://www.tinkutara.com/question/Q41838.png)
$${let}'{s}\:{say}\:\mathrm{0}<{b}\leqslant{a}\leqslant\mathrm{2}{R} \\ $$$$ \\ $$$${B}\left(\mathrm{0},\mathrm{0}\right) \\ $$$${O}\left(\frac{{b}}{\mathrm{2}},\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}\right) \\ $$$${C}\left({h},{r}\right) \\ $$$$ \\ $$$${Eqn}.\:{of}\:{BA}: \\ $$$${y}={mx}\:{with}\:{m}=\mathrm{tan}\:\left(\mathrm{cos}^{−\mathrm{1}} \frac{{b}}{\mathrm{2}{R}}\pm\mathrm{cos}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}{R}}\right) \\ $$$${Eqn}.\:{of}\:{circle}\:{with}\:{r}: \\ $$$$\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow\left({x}−{h}\right)+\left({y}−{r}\right)\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$${at}\:{M}: \\ $$$$\left({x}_{{M}} −{h}\right)+\left({y}_{{M}} −{r}\right){m}=\mathrm{0} \\ $$$$\left({x}_{{M}} −{h}\right)^{\mathrm{2}} +\left({y}_{{M}} −{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow\left({m}^{\mathrm{2}} +\mathrm{1}\right)\left({x}_{{M}} −{h}\right)^{\mathrm{2}} ={m}^{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$$\Rightarrow{x}_{{M}} ={h}\pm\frac{{mr}}{\:\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\Rightarrow{y}_{{M}} ={r}\mp\frac{{r}}{\:\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${y}_{{M}} ={mx}_{{M}} \\ $$$$\Rightarrow{y}_{{M}} ={r}\mp\frac{{r}}{\:\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}={mh}\pm\frac{{m}^{\mathrm{2}} {r}}{\:\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\Rightarrow\mathrm{1}\mp\frac{\mathrm{1}}{\:\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}}=\frac{{mh}}{{r}}\pm\frac{{m}^{\mathrm{2}} }{\:\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\Rightarrow{h}=\frac{{r}}{{m}}\left(\mathrm{1}\pm\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right)=\lambda{r}>\mathrm{0} \\ $$$${with}\:\lambda=\frac{\mathrm{1}}{{m}}\left(\mathrm{1}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$ \\ $$$${OC}={R}−{r} \\ $$$$\sqrt{\left({h}−\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({r}−\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}\right)^{\mathrm{2}} }={R}−{r} \\ $$$$\left(\lambda{r}−\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({r}−\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}\right)^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} \\ $$$$\lambda^{\mathrm{2}} {r}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{\mathrm{4}}−\lambda{br}+{r}^{\mathrm{2}} +{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{2}{r}\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}={R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{Rr} \\ $$$$\lambda^{\mathrm{2}} {r}=\lambda{b}+\mathrm{2}\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}−\mathrm{2}{R} \\ $$$$\Rightarrow{r}=\frac{\mathrm{1}}{\lambda^{\mathrm{2}} }\left[\lambda{b}−\mathrm{2}\left({R}−\sqrt{{R}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}\right)\right] \\ $$$${with}\:\lambda=\frac{\mathrm{1}}{{m}}\left(\mathrm{1}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$${and}\:{m}=\mathrm{tan}\:\left(\mathrm{cos}^{−\mathrm{1}} \frac{{b}}{\mathrm{2}{R}}\pm\mathrm{cos}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}{R}}\right) \\ $$$$=\frac{\sqrt{\mathrm{1}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} }}×\frac{{a}}{\mathrm{2}{R}}\pm\frac{{b}}{\mathrm{2}{R}}×\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} }}}{\frac{{b}}{\mathrm{2}{R}}×\frac{{a}}{\mathrm{2}{R}}\mp\sqrt{\mathrm{1}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} }}×\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} }}} \\ $$$$=\frac{{a}\sqrt{\mathrm{4}{R}^{\mathrm{2}} −{b}^{\mathrm{2}} }\pm{b}\sqrt{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{ab}\mp\sqrt{\left(\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\left(\mathrm{4}{R}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}} \\ $$$$ \\ $$$${with}\:\alpha=\frac{{a}}{\mathrm{2}{R}}\leqslant\mathrm{1},\:\beta=\frac{{b}}{\mathrm{2}{R}}\leqslant\mathrm{1}\:{and}\:\beta\leqslant\alpha, \\ $$$${m}=\frac{\alpha\sqrt{\mathrm{1}−\beta^{\mathrm{2}} }\pm\beta\sqrt{\mathrm{1}−\alpha^{\mathrm{2}} }}{\alpha\beta\mp\sqrt{\left(\mathrm{1}−\alpha^{\mathrm{2}} \right)\left(\mathrm{1}−\beta^{\mathrm{2}} \right)}} \\ $$$$\lambda=\frac{\mathrm{1}}{{m}}\left(\mathrm{1}+\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}{R}}{\lambda^{\mathrm{2}} }\left(\lambda\beta+\sqrt{\mathrm{1}−\beta^{\mathrm{2}} }−\mathrm{1}\right) \\ $$
Commented by ajfour last updated on 23/Aug/18
$${Thank}\:{you}\:{so}\:{much}\:{sir}\:! \\ $$
Commented by MrW3 last updated on 13/Aug/18
Commented by MrW3 last updated on 14/Aug/18
$${any}\:{other}\:{method}? \\ $$
Commented by ajfour last updated on 24/Aug/18
$${See}\:{Q}.\mathrm{42310}\:{Sir}.. \\ $$
Answered by ajfour last updated on 23/Aug/18
![r = 2R[((cos (((β−α)/2) ))/(cos (((α+β)/2) )))−tan (((α+β)/2) )]tan (((α+β)/2) ) with cos α=(a/(2R)) , cos β = (b/(2R)) . please help in checking this answer Sir!](https://www.tinkutara.com/question/Q42309.png)
$$\:\:{r}\:=\:\mathrm{2}{R}\left[\frac{\mathrm{cos}\:\left(\frac{\beta−\alpha}{\mathrm{2}}\:\right)}{\mathrm{cos}\:\left(\frac{\alpha+\beta}{\mathrm{2}}\:\right)}−\mathrm{tan}\:\left(\frac{\alpha+\beta}{\mathrm{2}}\:\right)\right]\mathrm{tan}\:\left(\frac{\alpha+\beta}{\mathrm{2}}\:\right)\: \\ $$$$\:\:\:\boldsymbol{{with}}\:\:\mathrm{cos}\:\alpha=\frac{{a}}{\mathrm{2}{R}}\:,\:\:\mathrm{cos}\:\beta\:=\:\frac{{b}}{\mathrm{2}{R}}\:. \\ $$$$ \\ $$$${please}\:{help}\:{in}\:{checking}\:{this}\:{answer} \\ $$$${Sir}! \\ $$