Question Number 175601 by ajfour last updated on 03/Sep/22
Answered by mr W last updated on 03/Sep/22
$${R}=\mathrm{1} \\ $$$${r}={radius}\:{of}\:{sphere} \\ $$$${if}\:{a}={b}: \\ $$$${r}\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }={aR} \\ $$$$\Rightarrow\frac{{r}}{{R}}=\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }} \\ $$$${if}\:{a}\neq{b}: \\ $$$$\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }−\frac{{ar}}{{R}}=\sqrt{{b}^{\mathrm{2}} +{R}^{\mathrm{2}} }−\frac{{br}}{{R}} \\ $$$$\Rightarrow\frac{{r}}{{R}}=\frac{\sqrt{{b}^{\mathrm{2}} +{R}^{\mathrm{2}} }−\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }}{{b}−{a}}=\frac{{a}+{b}}{\:\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} +{R}^{\mathrm{2}} }} \\ $$
Commented by ajfour last updated on 04/Sep/22
$${thanks}\:{for}\:{solving}\:{sir}. \\ $$
Commented by Tawa11 last updated on 15/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by ajfour last updated on 03/Sep/22
$${r}\left(\mathrm{sec}\:\theta+\mathrm{sec}\:\phi\right)={a}+{b} \\ $$$$\forall\:\:\mathrm{tan}\:\theta=\frac{{a}}{{R}},\:\mathrm{tan}\:\phi=\frac{{b}}{{R}} \\ $$$${r}=\frac{\left({a}+{b}\right){R}}{\:\sqrt{{R}^{\mathrm{2}} +{a}^{\mathrm{2}} }+\sqrt{{R}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$