Question Number 226372 by ajfour last updated on 26/Nov/25
Commented by ajfour last updated on 26/Nov/25
$${Find}\:{radii}\:\left({equal}\right)\:{of}\:{the}\:{two} \\ $$$${circles}.\:{Rectangle}\:{sides}\:{are} \\ $$$${length}\:{a}=\mathrm{2},\:{height}\:{b}=\mathrm{1}. \\ $$
Answered by fantastic2 last updated on 26/Nov/25
Commented by fantastic2 last updated on 26/Nov/25
$${a}−{r}+{b}−{r}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{r}=\frac{\mathrm{1}}{\mathrm{2}}\left(\left({a}+{b}\right)−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right) \\ $$
Commented by ajfour last updated on 26/Nov/25
$${very}\:{intelligent}! \\ $$
Commented by fantastic2 last updated on 26/Nov/25
$${another}\:{thing}\:{you}\:{can}\:{do} \\ $$$$\left({which}\:{is}\:{bit}\:{complex}\right) \\ $$$$\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}} \\ $$$$\frac{{r}}{{a}−{r}}=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$${r}\left(\mathrm{1}+\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\right)={a}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{{a}\mathrm{tan}\:\frac{\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}}}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}\:\frac{\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}}}{\mathrm{2}}} \\ $$
Answered by mr W last updated on 26/Nov/25
$$\left({a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right){r}={ab}=\mathrm{2}\: \\ $$$$\Rightarrow{r}=\frac{{ab}}{{a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:=\frac{{ab}\left({a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)}{\mathrm{2}{ab}} \\ $$$$\:\:\:\:\:\:\:=\frac{{a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$