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A-circle-tangents-to-x-and-y-axes-and-x-1-2-y-1-2-a-1-2-find-its-radious-




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

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