Question Number 49433 by behi83417@gmail.com last updated on 06/Dec/18
Commented by ajfour last updated on 06/Dec/18
$${circle}\:{radius},\:{Sir}\:? \\ $$
Commented by behi83417@gmail.com last updated on 06/Dec/18
$${BC}={x},{DC}={y},{circle}\:{radius}={r}\:. \\ $$$${find}\:{BD},{in}\:{terms}\:{of}:{x},{y},{r}\:\:. \\ $$
Commented by behi83417@gmail.com last updated on 07/Dec/18
![BD=a ∡BAD=α=BC^⌢ D ∡BCD=β=((BD^⌢ )/2)=((360−BC^⌢ D)/2)=180−((BC^⌢ D)/2) cosBC^ D=−cos((1/2)BA^ D) cos^2 β=((1+cosα)/2)⇒2cos^2 β−cosα=1 2(((x^2 +y^2 −a^2 )/(2xy)))^2 −((2r^2 −a^2 )/(2r^2 ))=1⇒ 4r^2 (x^2 +y^2 −a^2 )^2 −4x^2 y^2 (2r^2 −a^2 )=8r^2 x^2 y^2 4r^2 a^4 −4a^2 [2r^2 (x^2 +y^2 )−x^2 y^2 ]+4r^2 (x^2 +y^2 )^2 −16x^2 y^2 r^2 =0](https://www.tinkutara.com/question/Q49455.png)
$${BD}={a} \\ $$$$\measuredangle{BAD}=\alpha={B}\overset{\smallfrown} {{C}D} \\ $$$$\measuredangle{BCD}=\beta=\frac{{B}\overset{\smallfrown} {{D}}}{\mathrm{2}}=\frac{\mathrm{360}−{B}\overset{\smallfrown} {{C}D}}{\mathrm{2}}=\mathrm{180}−\frac{{B}\overset{\smallfrown} {{C}D}}{\mathrm{2}} \\ $$$${cosB}\overset{ } {{C}D}=−{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{B}\overset{ } {{A}D}\right) \\ $$$${cos}^{\mathrm{2}} \beta=\frac{\mathrm{1}+{cos}\alpha}{\mathrm{2}}\Rightarrow\mathrm{2}{cos}^{\mathrm{2}} \beta−{cos}\alpha=\mathrm{1} \\ $$$$\mathrm{2}\left(\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{xy}}\right)^{\mathrm{2}} −\frac{\mathrm{2}{r}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{r}^{\mathrm{2}} }=\mathrm{1}\Rightarrow \\ $$$$\mathrm{4}{r}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \left(\mathrm{2}{r}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)=\mathrm{8}{r}^{\mathrm{2}} {x}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$$\mathrm{4}{r}^{\mathrm{2}} {a}^{\mathrm{4}} −\mathrm{4}{a}^{\mathrm{2}} \left[\mathrm{2}{r}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−{x}^{\mathrm{2}} {y}^{\mathrm{2}} \right]+\mathrm{4}{r}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{16}{x}^{\mathrm{2}} {y}^{\mathrm{2}} {r}^{\mathrm{2}} =\mathrm{0} \\ $$
Commented by behi83417@gmail.com last updated on 07/Dec/18
$${you}\:{are}\:{right}\:{sir}.{this}\:{is}\:{my}\:{typo}.{now} \\ $$$${it}\:{is}\:{fixed}.{thanks}\:{a}\:{lot}\:{for}\:{attention}. \\ $$
Commented by mr W last updated on 07/Dec/18
![sir, but your eqn. differs from mine. yours is after rearrangement a^4 −[(x^2 +y^2 )−((x^2 y^2 )/r^2 )]a^2 +(x^2 −y^2 )^2 =0 mine a^4 −[2(x^2 +y^2 )−((x^2 y^2 )/r^2 )]a^2 +(x^2 −y^2 )^2 =0 can you please check?](https://www.tinkutara.com/question/Q49504.png)
$${sir},\:{but}\:{your}\:{eqn}.\:{differs}\:{from}\:{mine}. \\ $$$${yours}\:{is}\:{after}\:{rearrangement} \\ $$$${a}^{\mathrm{4}} −\left[\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\frac{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\right]{a}^{\mathrm{2}} +\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$${mine} \\ $$$${a}^{\mathrm{4}} −\left[\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\frac{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\right]{a}^{\mathrm{2}} +\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$${can}\:{you}\:{please}\:{check}? \\ $$
Commented by mr W last updated on 07/Dec/18
$${thanks}\:{for}\:{checking}\:{sir}! \\ $$
Answered by mr W last updated on 06/Dec/18
![let a=BD R=((xya)/( (√((x+y+a)(−x+y+a)(x−y+a)(x+y−a))))) R=((xya)/( (√([(x+y)^2 −a^2 ][a^2 −(x−y)^2 ])))) [(x+y)^2 −a^2 ][a^2 −(x−y)^2 ]=(((xy)/R))^2 a^2 a^4 −[2(x^2 +y^2 )−(((xy)/R))^2 ]a^2 +(x^2 −y^2 )^2 =0 a^2 =((2(x^2 +y^2 )−(((xy)/R))^2 ±(√([2(x^2 +y^2 )−(((xy)/R))^2 ]^2 −4(x^2 −y^2 )^2 )))/2) a^2 =((2(x^2 +y^2 )−(((xy)/R))^2 ±(√([4x^2 −(((xy)/R))^2 ][4y^2 −(((xy)/R))^2 ])))/2) a^2 =((2(x^2 +y^2 )−(((xy)/R))^2 ±(√((2x+((xy)/R))(2x−((xy)/R))(2y+((xy)/R))(2y−((xy)/R)))))/2) ⇒a=(√((2(x^2 +y^2 )−(((xy)/R))^2 ±(√((2x+((xy)/R))(2x−((xy)/R))(2y+((xy)/R))(2y−((xy)/R)))))/2))](https://www.tinkutara.com/question/Q49444.png)
$${let}\:{a}={BD} \\ $$$${R}=\frac{{xya}}{\:\sqrt{\left({x}+{y}+{a}\right)\left(−{x}+{y}+{a}\right)\left({x}−{y}+{a}\right)\left({x}+{y}−{a}\right)}} \\ $$$${R}=\frac{{xya}}{\:\sqrt{\left[\left({x}+{y}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \right]\left[{a}^{\mathrm{2}} −\left({x}−{y}\right)^{\mathrm{2}} \right]}} \\ $$$$\left[\left({x}+{y}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \right]\left[{a}^{\mathrm{2}} −\left({x}−{y}\right)^{\mathrm{2}} \right]=\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} {a}^{\mathrm{2}} \\ $$$${a}^{\mathrm{4}} −\left[\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \right]{a}^{\mathrm{2}} +\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \pm\sqrt{\left[\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \right]^{\mathrm{2}} −\mathrm{4}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \pm\sqrt{\left[\mathrm{4}{x}^{\mathrm{2}} −\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \right]\left[\mathrm{4}{y}^{\mathrm{2}} −\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \right]}}{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \pm\sqrt{\left(\mathrm{2}{x}+\frac{{xy}}{{R}}\right)\left(\mathrm{2}{x}−\frac{{xy}}{{R}}\right)\left(\mathrm{2}{y}+\frac{{xy}}{{R}}\right)\left(\mathrm{2}{y}−\frac{{xy}}{{R}}\right)}}{\mathrm{2}} \\ $$$$\Rightarrow{a}=\sqrt{\frac{\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \pm\sqrt{\left(\mathrm{2}{x}+\frac{{xy}}{{R}}\right)\left(\mathrm{2}{x}−\frac{{xy}}{{R}}\right)\left(\mathrm{2}{y}+\frac{{xy}}{{R}}\right)\left(\mathrm{2}{y}−\frac{{xy}}{{R}}\right)}}{\mathrm{2}}} \\ $$
Commented by behi83417@gmail.com last updated on 06/Dec/18
$${wow}!\:{perfect}\:{sir}.{thanks}\:{in}\:{advance}. \\ $$
Commented by mr W last updated on 07/Dec/18
$${check}\:{with}\:{x}={y}=\sqrt{\mathrm{2}}{R} \\ $$$$\Rightarrow{a}=\sqrt{\frac{\mathrm{2}\left(\mathrm{4}{R}^{\mathrm{2}} \right)−\left(\mathrm{2}{R}\right)^{\mathrm{2}} \pm\sqrt{\left(\mathrm{2}\sqrt{\mathrm{2}}{R}+\mathrm{2}{R}\right)^{\mathrm{2}} \left(\mathrm{2}\sqrt{\mathrm{2}}{R}−\mathrm{2}{R}\right)^{\mathrm{2}} }}{\mathrm{2}}} \\ $$$$\Rightarrow{a}=\sqrt{\frac{\mathrm{4}{R}^{\mathrm{2}} \pm\mathrm{4}{R}^{\mathrm{2}} }{\mathrm{2}}}=\mathrm{2}{R}\:{or}\:\mathrm{0}\Rightarrow{correct} \\ $$
Answered by behi83417@gmail.com last updated on 06/Dec/18
Commented by behi83417@gmail.com last updated on 06/Dec/18
$$\measuredangle{FBC}=\measuredangle{FDC}=\mathrm{90}^{\bullet} \\ $$$${BF}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} ,{DF}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$${by}\:{using}\:\boldsymbol{\mathrm{petolemy}}'\boldsymbol{\mathrm{s}}\:{teorem}\:{in}:{FDCB}: \\ $$$${BD}.{CF}={BC}.{DF}+{BF}.{CD} \\ $$$$\mathrm{2}{r}.{BD}={x}.\sqrt{\mathrm{4}{r}^{\mathrm{2}} −{y}^{\mathrm{2}} }+{y}.\sqrt{\mathrm{4}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{BD}=\frac{\boldsymbol{\mathrm{x}}.\sqrt{\mathrm{4}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} }+\boldsymbol{\mathrm{y}}.\sqrt{\mathrm{4}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}{\mathrm{2}\boldsymbol{\mathrm{r}}}\:.\blacksquare \\ $$
Commented by behi83417@gmail.com last updated on 07/Dec/18
![chech for:x=y=r(√2) BD=((r(√2).r(√2)+r(√2).r(√2))/(2r))=((4r^2 )/(2r))=2r [ok].](https://www.tinkutara.com/question/Q49524.png)
$${chech}\:{for}:{x}={y}={r}\sqrt{\mathrm{2}} \\ $$$${BD}=\frac{{r}\sqrt{\mathrm{2}}.{r}\sqrt{\mathrm{2}}+{r}\sqrt{\mathrm{2}}.{r}\sqrt{\mathrm{2}}}{\mathrm{2}{r}}=\frac{\mathrm{4}{r}^{\mathrm{2}} }{\mathrm{2}{r}}=\mathrm{2}{r}\:\left[{ok}\right]. \\ $$
Answered by behi83417@gmail.com last updated on 07/Dec/18
$$\measuredangle{BAC}=\alpha,\measuredangle{DAC}=\beta \\ $$$${cos}\alpha=\frac{\mathrm{2}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }{\mathrm{2}{r}^{\mathrm{2}} },{sin}\alpha=\frac{{x}}{\mathrm{2}{r}^{\mathrm{2}} }\sqrt{\mathrm{4}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$${cos}\beta=\frac{\mathrm{2}{r}^{\mathrm{2}} −{y}^{\mathrm{2}} }{\mathrm{2}{r}^{\mathrm{2}} },{sin}\beta=\frac{{y}}{\mathrm{2}{r}^{\mathrm{2}} }\sqrt{\mathrm{4}{r}^{\mathrm{2}} −{y}^{\mathrm{2}} } \\ $$$${cos}\left(\alpha+\beta\right)=\frac{\left(\mathrm{2}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)\left(\mathrm{2}{r}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}{\mathrm{4}{r}^{\mathrm{4}} }−\frac{{xy}}{\mathrm{4}{r}^{\mathrm{4}} }\sqrt{\left(\mathrm{4}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)\left(\mathrm{4}{r}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)} \\ $$$${BD}^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} .{cos}\left(\alpha+\beta\right)= \\ $$$$=\frac{\boldsymbol{\mathrm{xy}}.\sqrt{\left(\mathrm{4}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\left(\mathrm{4}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right)}−\left(\mathrm{2}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\left(\mathrm{2}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right)+\mathrm{4}\boldsymbol{\mathrm{r}}^{\mathrm{4}} }{\mathrm{2}\boldsymbol{\mathrm{r}}^{\mathrm{2}} }. \\ $$$$=\frac{\boldsymbol{\mathrm{xy}}.\sqrt{\left(\mathrm{4}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\left(\mathrm{4}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right)}+\mathrm{2}{r}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{\mathrm{2}\boldsymbol{\mathrm{r}}^{\mathrm{2}} }. \\ $$$${BD}=\sqrt{\frac{{xy}}{\mathrm{2}{r}^{\mathrm{2}} }.\sqrt{\left(\mathrm{4}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)\left(\mathrm{4}{r}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}+\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left(\frac{{xy}}{{r}\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }. \\ $$