Question Number 42310 by ajfour last updated on 23/Aug/18
Commented by ajfour last updated on 23/Aug/18
$${Solution}\:{to}\:{Q}.\mathrm{41766}\:\left({Another}\:{method}\right) \\ $$
Answered by ajfour last updated on 23/Aug/18
![cos α = (a/(2R)) ; cos β = (b/(2R)) θ=((α+β)/2) eq. of bigger circle : x^2 +y^2 =R^( 2) ...(i) eq. of small circle : (x−h)^2 +(y−k)^2 =r^2 ...(ii) with h^2 +k^2 =(R−r)^2 ...(iii) but h=(r/(sin θ)) cos (β−θ)−R and k =− (r/(sin θ)) sin (β−θ) substituting these in (iii) we have (r^2 /(sin^2 θ))−((2rR)/(sin θ)) cos (β−θ)=r^2 −2rR ⇒ r=0 or r = ((2R[((cos (β−θ))/(sin θ))−1])/((1/(sin^2 θ))−1)) but θ= ((α+β)/2) , hence r = 2R[((cos (((β−α)/2)))/(cos (((β+α)/2))))−tan (((β+α)/2))]tan (((β+α)/2)) ; with β = cos^(−1) (b/(2R)) & α=cos^(−1) (a/(2R)) .](Q42312.png)
$$\mathrm{cos}\:\alpha\:=\:\frac{{a}}{\mathrm{2}{R}}\:;\:\:\mathrm{cos}\:\beta\:=\:\frac{{b}}{\mathrm{2}{R}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\theta=\frac{\alpha+\beta}{\mathrm{2}} \\ $$$${eq}.\:{of}\:{bigger}\:{circle}\:: \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={R}^{\:\mathrm{2}} \:\:\:\:\:\:...\left({i}\right) \\ $$$${eq}.\:{of}\:{small}\:{circle}\:: \\ $$$$\:\:\:\:\:\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \:\:\:\:...\left({ii}\right) \\ $$$${with}\:\:\:\:\boldsymbol{{h}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} =\left(\boldsymbol{{R}}−\boldsymbol{{r}}\right)^{\mathrm{2}} \:\:\:\:\:...\left({iii}\right) \\ $$$$\:\:\:{but}\:\:\:\boldsymbol{{h}}=\frac{\boldsymbol{{r}}}{\mathrm{sin}\:\theta}\:\mathrm{cos}\:\left(\beta−\theta\right)−{R} \\ $$$${and}\:\:\:\:\boldsymbol{{k}}\:=−\:\frac{\boldsymbol{{r}}}{\mathrm{sin}\:\theta}\:\mathrm{sin}\:\left(\beta−\theta\right) \\ $$$${substituting}\:{these}\:{in}\:\left({iii}\right)\:{we}\:{have} \\ $$$$\:\:\frac{{r}^{\mathrm{2}} }{\mathrm{sin}\:^{\mathrm{2}} \theta}−\frac{\mathrm{2}{rR}}{\mathrm{sin}\:\theta}\:\mathrm{cos}\:\left(\beta−\theta\right)={r}^{\mathrm{2}} −\mathrm{2}{rR} \\ $$$$\Rightarrow\:\:\:{r}=\mathrm{0}\:\:{or} \\ $$$$\:\:\:\boldsymbol{{r}}\:=\:\frac{\mathrm{2}{R}\left[\frac{\mathrm{cos}\:\left(\beta−\theta\right)}{\mathrm{sin}\:\theta}−\mathrm{1}\right]}{\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \theta}−\mathrm{1}} \\ $$$${but}\:\:\:\theta=\:\frac{\alpha+\beta}{\mathrm{2}}\:\:\:,\:{hence} \\ $$$$\boldsymbol{{r}}\:=\:\mathrm{2}\boldsymbol{{R}}\left[\frac{\mathrm{cos}\:\left(\frac{\beta−\alpha}{\mathrm{2}}\right)}{\mathrm{cos}\:\left(\frac{\beta+\alpha}{\mathrm{2}}\right)}−\mathrm{tan}\:\left(\frac{\beta+\alpha}{\mathrm{2}}\right)\right]\mathrm{tan}\:\left(\frac{\beta+\alpha}{\mathrm{2}}\right)\:; \\ $$$${with}\:\:\beta\:=\:\mathrm{cos}^{−\mathrm{1}} \frac{{b}}{\mathrm{2}{R}}\:\:\&\:\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}{R}}\:. \\ $$
Commented by MrW3 last updated on 24/Aug/18
$${very}\:{nice}!\:{the}\:{answer}\:{is}\:{correct}. \\ $$