Question-226375

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Question Number 226375 by ajfour last updated on 26/Nov/25

Commented by ajfour last updated on 26/Nov/25

$${Semicircle}\:{radius}\:{is}\:{R}.\:{Square} \\ $$$${side}\:{is}\:{s}.\:{Find}\:{circle}\:{radius}\:{r}. \\ $$

Answered by fantastic2 last updated on 26/Nov/25

Commented by fantastic2 last updated on 26/Nov/25

(r−((√(R^2 −s^2 ))−s))^2 +r^2 =(R−r)^2 r^2 −2r((√(R^2 −s^2 ))−s)+((√(R^2 −s^2 ))−s)^2 +r^2 =R^2 +r^2 −2rR r^2 −2r((√(R^2 −s^2 ))−s+R)=R^2 −((√(R^2 −s^2 ))−s)^2 r^2 −2r(x+R)=R^2 −x^2 r^2 −2r(x+R)+(x+R)(x−R)=0 r=((2(x+R)±(√(2(x+R)^2 −4(x+R)(x−R))))/2)

$$\left({r}−\left(\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} }−{s}\right)\right)^{\mathrm{2}} +{r}^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{r}\left(\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} }−{s}\right)+\left(\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} }−{s}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} ={R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{rR} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{r}\left(\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} }−{s}+{R}\right)={R}^{\mathrm{2}} −\left(\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} }−{s}\right)^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{r}\left({x}+{R}\right)={R}^{\mathrm{2}} −{x}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{r}\left({x}+{R}\right)+\left({x}+{R}\right)\left({x}−{R}\right)=\mathrm{0} \\ $$$${r}=\frac{\mathrm{2}\left({x}+{R}\right)\pm\sqrt{\mathrm{2}\left({x}+{R}\right)^{\mathrm{2}} −\mathrm{4}\left({x}+{R}\right)\left({x}−{R}\right)}}{\mathrm{2}} \\ $$$$ \\ $$

Answered by mr W last updated on 27/Nov/25

Commented by mr W last updated on 27/Nov/25

case 1: let λ=(s/R), ξ=(r/R) x_A =s−(√(R^2 −s^2 )) y_A =s C(h,r) h^2 =(R−r)^2 −r^2 =R^2 −2Rr ⇒r=((R^2 −h^2 )/(2R)) (h−x_A )^2 +(r−s)^2 =r^2 h^2 −2hx_A +x_A ^2 −2sr+s^2 =0 h^2 −2h(s−(√(R^2 −s^2 )))+s^2 +R^2 −s^2 −2s(√(R^2 −s^2 ))−2sr+s^2 =0 (1+λ)((h/R))^2 −2(λ−(√(1−λ^2 )))((h/R))+1+λ^2 −λ−2λ(√(1−λ^2 ))=0 ⇒(h/R)=((λ+λ(√(2(√(1−λ^2 ))−λ))−(√(1−λ^2 )))/(1+λ)) ⇒(r/R)=(1/2)(1−(h^2 /R^2 ))

$${case}\:\mathrm{1}: \\ $$$${let}\:\lambda=\frac{{s}}{{R}},\:\xi=\frac{{r}}{{R}} \\ $$$${x}_{{A}} ={s}−\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} } \\ $$$${y}_{{A}} ={s} \\ $$$${C}\left({h},{r}\right) \\ $$$${h}^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} ={R}^{\mathrm{2}} −\mathrm{2}{Rr} \\ $$$$\Rightarrow{r}=\frac{{R}^{\mathrm{2}} −{h}^{\mathrm{2}} }{\mathrm{2}{R}} \\ $$$$\left({h}−{x}_{{A}} \right)^{\mathrm{2}} +\left({r}−{s}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${h}^{\mathrm{2}} −\mathrm{2}{hx}_{{A}} +{x}_{{A}} ^{\mathrm{2}} −\mathrm{2}{sr}+{s}^{\mathrm{2}} =\mathrm{0} \\ $$$${h}^{\mathrm{2}} −\mathrm{2}{h}\left({s}−\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} }\right)+{s}^{\mathrm{2}} +{R}^{\mathrm{2}} −{s}^{\mathrm{2}} −\mathrm{2}{s}\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} }−\mathrm{2}{sr}+{s}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{1}+\lambda\right)\left(\frac{{h}}{{R}}\right)^{\mathrm{2}} −\mathrm{2}\left(\lambda−\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }\right)\left(\frac{{h}}{{R}}\right)+\mathrm{1}+\lambda^{\mathrm{2}} −\lambda−\mathrm{2}\lambda\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\frac{{h}}{{R}}=\frac{\lambda+\lambda\sqrt{\mathrm{2}\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }−\lambda}−\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }}{\mathrm{1}+\lambda} \\ $$$$\Rightarrow\frac{{r}}{{R}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{{h}^{\mathrm{2}} }{{R}^{\mathrm{2}} }\right) \\ $$

Commented by mr W last updated on 27/Nov/25

Commented by mr W last updated on 27/Nov/25

Commented by mr W last updated on 27/Nov/25

case 2: let λ=(s/R), ξ=(r/R) x_B =s−(√(R^2 −s^2 )) C(h,r) h^2 =(R−r)^2 −r^2 =R^2 −2Rr ⇒r=((R^2 −h^2 )/(2R)) x_B +r=h s−(√(R^2 −s^2 ))+((R^2 −h^2 )/(2R))=h 2λ−2(√(1−λ^2 ))+1−((h/R))^2 =2((h/R)) ((h/R))^2 +2((h/R))+2(√(1−λ^2 ))−1−2λ=0 ⇒(h/R)=−1+(√(2(1+λ−(√(1−λ^2 ))))) ⇒(r/R)=(1/2)(1−(h^2 /R^2 )) (√(((s/2))^2 +s^2 ))≤R ⇒(s/R)≤(2/( (√5)))≈0.894 r=s: ⇒(h^2 /R^2 )=1−2λ ⇒(√(1−2λ))=−1+(√(2(1+λ−(√(1−λ^2 ))))) ⇒λ≈0.4932 i.e. if (s/R)≤0.4932 ⇒case 1 if 0.4932≤(s/R) ≤(2/( (√5))) ⇒case 2

$${case}\:\mathrm{2}: \\ $$$${let}\:\lambda=\frac{{s}}{{R}},\:\xi=\frac{{r}}{{R}} \\ $$$${x}_{{B}} ={s}−\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} } \\ $$$${C}\left({h},{r}\right) \\ $$$${h}^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} ={R}^{\mathrm{2}} −\mathrm{2}{Rr} \\ $$$$\Rightarrow{r}=\frac{{R}^{\mathrm{2}} −{h}^{\mathrm{2}} }{\mathrm{2}{R}} \\ $$$${x}_{{B}} +{r}={h} \\ $$$${s}−\sqrt{{R}^{\mathrm{2}} −{s}^{\mathrm{2}} }+\frac{{R}^{\mathrm{2}} −{h}^{\mathrm{2}} }{\mathrm{2}{R}}={h} \\ $$$$\mathrm{2}\lambda−\mathrm{2}\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }+\mathrm{1}−\left(\frac{{h}}{{R}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{{h}}{{R}}\right) \\ $$$$\left(\frac{{h}}{{R}}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{{h}}{{R}}\right)+\mathrm{2}\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }−\mathrm{1}−\mathrm{2}\lambda=\mathrm{0} \\ $$$$\Rightarrow\frac{{h}}{{R}}=−\mathrm{1}+\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda−\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }\right)} \\ $$$$\Rightarrow\frac{{r}}{{R}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{{h}^{\mathrm{2}} }{{R}^{\mathrm{2}} }\right) \\ $$$$ \\ $$$$\sqrt{\left(\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} +{s}^{\mathrm{2}} }\leqslant{R} \\ $$$$\Rightarrow\frac{{s}}{{R}}\leqslant\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\approx\mathrm{0}.\mathrm{894} \\ $$$${r}={s}: \\ $$$$\Rightarrow\frac{{h}^{\mathrm{2}} }{{R}^{\mathrm{2}} }=\mathrm{1}−\mathrm{2}\lambda \\ $$$$\Rightarrow\sqrt{\mathrm{1}−\mathrm{2}\lambda}=−\mathrm{1}+\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda−\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }\right)} \\ $$$$\Rightarrow\lambda\approx\mathrm{0}.\mathrm{4932} \\ $$$${i}.{e}.\:{if}\:\frac{{s}}{{R}}\leqslant\mathrm{0}.\mathrm{4932}\:\Rightarrow{case}\:\mathrm{1} \\ $$$${if}\:\:\mathrm{0}.\mathrm{4932}\leqslant\frac{{s}}{{R}}\:\leqslant\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow{case}\:\mathrm{2} \\ $$

Commented by mr W last updated on 27/Nov/25