Coordinate Geometry Questions

Question Number 102060 by ajfour last updated on 06/Jul/20

Commented by ajfour last updated on 06/Jul/20

$${Find}\:\boldsymbol{{r}}\:{and}\:\boldsymbol{{R}}. \\$$

Answered by mr W last updated on 06/Jul/20

$${center}\:{of}\:{circle}\:{r}\:{inside}\:{M}\left(\mathrm{0},{m}\right) \\$$$${m}=\frac{\mathrm{1}}{\mathrm{4}}+{r}^{\mathrm{2}} \\$$$${center}\:{of}\:{circle}\:{r}\:{outside}\:{N}\left(\sqrt{\mathrm{4}{r}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{4}}+{r}^{\mathrm{2}} −{r}\right)^{\mathrm{2}} },{r}\right) \\$$$$\frac{\mathrm{1}}{\mathrm{2}}\left({r}+\frac{\mathrm{1}}{\mathrm{4}}+{r}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{4}{r}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{4}}+{r}^{\mathrm{2}} −{r}\right)^{\mathrm{2}} \right] \\$$$$\mathrm{2}{r}^{\mathrm{2}} −\mathrm{2}{r}−\frac{\mathrm{1}}{\mathrm{2}}=\left(\frac{\mathrm{1}}{\mathrm{4}}+{r}^{\mathrm{2}} −{r}\right)^{\mathrm{2}} \\$$$${r}^{\mathrm{4}} −\mathrm{2}{r}^{\mathrm{3}} −\frac{{r}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{3}{r}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{16}}=\mathrm{0} \\$$$$\left({r}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \left({r}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{0} \\$$$$\Rightarrow{r}=\frac{\mathrm{3}}{\mathrm{2}} \\$$$$\\$$$${center}\:{of}\:{circle}\:{R}\:{at}\:\left({R},{h}\right) \\$$$${R}^{\mathrm{2}} +\left({h}−\frac{\mathrm{1}}{\mathrm{4}}−{r}^{\mathrm{2}} \right)^{\mathrm{2}} =\left({R}+{r}\right)^{\mathrm{2}} \:\:\:...\left({i}\right) \\$$$$\left(\sqrt{\mathrm{4}{r}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{4}}+{r}^{\mathrm{2}} −{r}\right)^{\mathrm{2}} }−{R}\right)^{\mathrm{2}} +\left({h}−{r}\right)^{\mathrm{2}} =\left({R}+{r}\right)^{\mathrm{2}} \:\:\:...\left({ii}\right) \\$$$$\left({i}\right)−\left({ii}\right): \\$$$$\left(\mathrm{2}{R}−\sqrt{\mathrm{4}{r}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{4}}+{r}^{\mathrm{2}} −{r}\right)^{\mathrm{2}} }\right)\sqrt{\mathrm{4}{r}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{4}}+{r}^{\mathrm{2}} −{r}\right)^{\mathrm{2}} }+\left(\mathrm{2}{h}−\frac{\mathrm{1}}{\mathrm{4}}−{r}^{\mathrm{2}} −{r}\right)\left(−\frac{\mathrm{1}}{\mathrm{4}}−{r}^{\mathrm{2}} +{r}\right)=\mathrm{0} \\$$$$\Rightarrow{h}=\mathrm{2}\left(\sqrt{\mathrm{2}}{R}−\mathrm{1}\right) \\$$$${R}^{\mathrm{2}} +\left(\mathrm{2}\sqrt{\mathrm{2}}{R}−\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} =\left({R}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \\$$$$\mathrm{8}{R}^{\mathrm{2}} −\mathrm{3}\left(\mathrm{1}+\mathrm{6}\sqrt{\mathrm{2}}\right){R}+\mathrm{18}=\mathrm{0} \\$$$$\Rightarrow{oR}=\frac{\mathrm{3}}{\mathrm{16}}\left[\mathrm{1}+\mathrm{6}\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}\left(\mathrm{3}+\mathrm{4}\sqrt{\mathrm{2}}\right)}\right] \\$$

Commented by mr W last updated on 06/Jul/20

Commented by ajfour last updated on 07/Jul/20

$${Thank}\:{you}\:{Sir},\:{Excellent}! \\$$

Answered by ajfour last updated on 07/Jul/20

$${let}\:{the}\:{smaller}\:{circles}\:{touch}\:{parabola} \\$$$${at}\:{P}\left({h},{h}^{\mathrm{2}} \right) \\$$$${h}={r}\mathrm{sin}\:\theta\:\:\:,\:\:\:\:{h}^{\mathrm{2}} ={r}+{r}\mathrm{cos}\:\theta \\$$$${And}\:\:\:\:\:\mathrm{2}{h}=\mathrm{tan}\:\theta \\$$$$\Rightarrow\:\:\:\mathrm{tan}\:\theta=\mathrm{2}{r}\mathrm{sin}\:\theta\:\:\:\Rightarrow\:\:\:\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}{r}} \\$$$$\&\:\:\:\:\:{r}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{r}^{\mathrm{2}} }\right)={r}+\frac{\mathrm{1}}{\mathrm{2}} \\$$$$\Rightarrow\:\:\:\:{r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}={r}+\frac{\mathrm{1}}{\mathrm{2}} \\$$$$\:\:\:\:\:\:\:\:{r}^{\mathrm{2}} −{r}−\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{0}\:\:\:\:\Rightarrow\:\:{r}=\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}} \\$$$$\:\:\:\:{r}=\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:,\:\:\:\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{3}} \\$$$${center}\:{of}\:{left}\:{smaller}\:{circle}\:{be}\:{C}. \\$$$${y}_{{C}} =\:{r}+\mathrm{2}{r}\mathrm{cos}\:\theta\:=\:\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\$$$$\:\:{Altitude}\:{of}\:{isosceles}\:{triangle} \\$$$$\:\:\:\:\:{H}=\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\$$$${H}\mathrm{cos}\:\theta+{r}\mathrm{sin}\:\theta={R} \\$$$$\Rightarrow\:\:\frac{\mathrm{1}}{\mathrm{9}}\left({R}^{\mathrm{2}} +\mathrm{2}{Rr}\right)=\left({R}−\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}\right)^{\mathrm{2}} \\$$$$\Rightarrow\:\:\:{R}^{\mathrm{2}} +\mathrm{3}{R}=\mathrm{9}\left[{R}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}{R}+\mathrm{2}\right] \\$$$$\Rightarrow\:\:\:\:\mathrm{8}{R}^{\mathrm{2}} −\left(\mathrm{18}\sqrt{\mathrm{2}}+\mathrm{3}\right){R}+\mathrm{18}=\mathrm{0} \\$$$$\:\:\:{R}=\frac{\mathrm{18}\sqrt{\mathrm{2}}+\mathrm{3}}{\mathrm{16}}+\sqrt{\left(\frac{\mathrm{18}\sqrt{\mathrm{2}}+\mathrm{3}}{\mathrm{16}}\right)^{\mathrm{2}} −\frac{\mathrm{18}}{\mathrm{8}}} \\$$$$\:\:\:{R}\:=\:\frac{\mathrm{3}}{\mathrm{16}}\left(\mathrm{6}\sqrt{\mathrm{2}}+\mathrm{1}+\sqrt{\mathrm{9}+\mathrm{12}\sqrt{\mathrm{2}}}\right)\:. \\$$