Question Number 213573 by mr W last updated on 09/Nov/24
Commented by mr W last updated on 09/Nov/24
$${find}\:{r}=? \\ $$
Commented by ajfour last updated on 09/Nov/24
Sir, this we hsd solved before.
Commented by mr W last updated on 09/Nov/24
$${i}'{ve}\:{forgotten}\:{sir}. \\ $$
Answered by mr W last updated on 09/Nov/24
Commented by mr W last updated on 09/Nov/24
$${R}=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\alpha}}{\mathrm{2}\:\mathrm{sin}\:\alpha} \\ $$$${P}\left(\frac{{a}}{\mathrm{2}},\:\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\right) \\ $$$${Q}\left(\frac{{r}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}},\:{r}\right) \\ $$$${PQ}={R}−{r} \\ $$$$\sqrt{\left(\frac{{a}}{\mathrm{2}}−\frac{{r}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}\right)^{\mathrm{2}} +\left(\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}−{r}\right)^{\mathrm{2}} }={R}−{r} \\ $$$$\frac{{r}}{\mathrm{tan}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}}=\frac{{a}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}+\mathrm{2}\left(\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}−{R}\right) \\ $$$$\Rightarrow{r}={a}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}+\mathrm{2}\left(\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}−{R}\right)\mathrm{tan}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 10/Nov/24
$${If}\:{smaller}\:{circle}\:{is}\:{small},\:{then} \\ $$$$\left({R}−{r}\right)^{\mathrm{2}} =\left(\frac{{r}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({r}+\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{r}=\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)\left\{\frac{{a}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}−\mathrm{2}}−\mathrm{2}{R}−\mathrm{2}\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\:\right\} \\ $$