Question Number 58282 by behi83417@gmail.com last updated on 20/Apr/19
$$\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{circle}}\:\boldsymbol{\mathrm{tangents}}\:\boldsymbol{\mathrm{to}}\::\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}}\:\boldsymbol{\mathrm{axes}}\:\boldsymbol{\mathrm{and}} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{2}}} +\boldsymbol{\mathrm{y}}^{\frac{\mathrm{1}}{\mathrm{2}}} =\boldsymbol{\mathrm{a}}^{\frac{\mathrm{1}}{\mathrm{2}}} .\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{its}}\:\boldsymbol{\mathrm{radious}}. \\ $$
Answered by mr W last updated on 21/Apr/19
$${due}\:{to}\:{symmetry}\:{the}\:{circle}\:{touches} \\ $$$${the}\:{curve}\:{at}\:\left({t},{t}\right). \\ $$$$\sqrt{{t}}+\sqrt{{t}}=\sqrt{{a}} \\ $$$$\Rightarrow{t}=\frac{{a}}{\mathrm{4}} \\ $$$${eqn}.\:{of}\:{circle}:\:\left({x}−{r}\right)^{\mathrm{2}} +\left({y}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left({t}−{r}\right)^{\mathrm{2}} +\left({t}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{2}}\left({t}−{r}\right)=\pm{r} \\ $$$$\Rightarrow{r}=\frac{\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}}+\mathrm{1}}{t}=\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){t}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{4}}{a} \\ $$$${or} \\ $$$$\Rightarrow{r}=\frac{\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}}−\mathrm{1}}{t}=\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){t}=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{4}}{a} \\ $$$$\left({but}\:{this}\:{circle}\:{intersects}\:{also}\:{with}\right. \\ $$$$\left.{the}\:{curve}.\right) \\ $$
Commented by behi83417@gmail.com last updated on 21/Apr/19
$${thanks}\:{a}\:{lot}\:{dear}\:{master}. \\ $$$${i}\:{have}\:{some}\:{problem}\:{with}\:{last}\:{line}! \\ $$
Commented by mr W last updated on 21/Apr/19
$${the}\:{circle}\:{with}\:{r}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{4}}{a}\:{only}\:{tangents} \\ $$$${the}\:{curve},\:{but}\:{the}\:{circle}\:{with}\:{r}=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{4}}{a} \\ $$$${not}\:{only}\:{tangents}\:{the}\:{curve}\:{but}\:{also} \\ $$$${intersects}\:{with}\:{it}. \\ $$
Commented by mr W last updated on 21/Apr/19
Commented by behi83417@gmail.com last updated on 21/Apr/19
$${thank}\:{you}\:{so}\:{much}\:{master}. \\ $$$${perfect}!{now}\:{it}\:{is}\:{clear}\:{to}\:{me}. \\ $$