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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 ?

$${circle}\:{radius},\:{Sir}\:? \\ $$

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  .

$${BC}={x},{DC}={y},{circle}\:{radius}={r}\:. \\ $$$${find}\:{BD},{in}\:{terms}\:{of}:{x},{y},{r}\:\:. \\ $$

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

$${BD}={a} \\ $$$$\measuredangle{BAD}=\alpha={B}\overset{\smallfrown} {{C}D} \\ $$$$\measuredangle{BCD}=\beta=\frac{{B}\overset{\smallfrown} {{D}}}{\mathrm{2}}=\frac{\mathrm{360}−{B}\overset{\smallfrown} {{C}D}}{\mathrm{2}}=\mathrm{180}−\frac{{B}\overset{\smallfrown} {{C}D}}{\mathrm{2}} \\ $$$${cosB}\overset{} {{C}D}=−{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{B}\overset{} {{A}D}\right) \\ $$$${cos}^{\mathrm{2}} \beta=\frac{\mathrm{1}+{cos}\alpha}{\mathrm{2}}\Rightarrow\mathrm{2}{cos}^{\mathrm{2}} \beta−{cos}\alpha=\mathrm{1} \\ $$$$\mathrm{2}\left(\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{xy}}\right)^{\mathrm{2}} −\frac{\mathrm{2}{r}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{r}^{\mathrm{2}} }=\mathrm{1}\Rightarrow \\ $$$$\mathrm{4}{r}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \left(\mathrm{2}{r}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)=\mathrm{8}{r}^{\mathrm{2}} {x}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$$\mathrm{4}{r}^{\mathrm{2}} {a}^{\mathrm{4}} −\mathrm{4}{a}^{\mathrm{2}} \left[\mathrm{2}{r}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−{x}^{\mathrm{2}} {y}^{\mathrm{2}} \right]+\mathrm{4}{r}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{16}{x}^{\mathrm{2}} {y}^{\mathrm{2}} {r}^{\mathrm{2}} =\mathrm{0} \\ $$

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.

$${you}\:{are}\:{right}\:{sir}.{this}\:{is}\:{my}\:{typo}.{now} \\ $$$${it}\:{is}\:{fixed}.{thanks}\:{a}\:{lot}\:{for}\:{attention}. \\ $$

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?

$${sir},\:{but}\:{your}\:{eqn}.\:{differs}\:{from}\:{mine}. \\ $$$${yours}\:{is}\:{after}\:{rearrangement} \\ $$$${a}^{\mathrm{4}} −\left[\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\frac{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\right]{a}^{\mathrm{2}} +\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$${mine} \\ $$$${a}^{\mathrm{4}} −\left[\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\frac{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\right]{a}^{\mathrm{2}} +\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$${can}\:{you}\:{please}\:{check}? \\ $$

Commented by mr W last updated on 07/Dec/18

thanks for checking sir!

$${thanks}\:{for}\:{checking}\:{sir}! \\ $$

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))

$${let}\:{a}={BD} \\ $$$${R}=\frac{{xya}}{\sqrt{\left({x}+{y}+{a}\right)\left(−{x}+{y}+{a}\right)\left({x}−{y}+{a}\right)\left({x}+{y}−{a}\right)}} \\ $$$${R}=\frac{{xya}}{\sqrt{\left[\left({x}+{y}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \right]\left[{a}^{\mathrm{2}} −\left({x}−{y}\right)^{\mathrm{2}} \right]}} \\ $$$$\left[\left({x}+{y}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \right]\left[{a}^{\mathrm{2}} −\left({x}−{y}\right)^{\mathrm{2}} \right]=\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} {a}^{\mathrm{2}} \\ $$$${a}^{\mathrm{4}} −\left[\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \right]{a}^{\mathrm{2}} +\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \pm\sqrt{\left[\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \right]^{\mathrm{2}} −\mathrm{4}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \pm\sqrt{\left[\mathrm{4}{x}^{\mathrm{2}} −\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \right]\left[\mathrm{4}{y}^{\mathrm{2}} −\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \right]}}{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \pm\sqrt{\left(\mathrm{2}{x}+\frac{{xy}}{{R}}\right)\left(\mathrm{2}{x}−\frac{{xy}}{{R}}\right)\left(\mathrm{2}{y}+\frac{{xy}}{{R}}\right)\left(\mathrm{2}{y}−\frac{{xy}}{{R}}\right)}}{\mathrm{2}} \\ $$$$\Rightarrow{a}=\sqrt{\frac{\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left(\frac{{xy}}{{R}}\right)^{\mathrm{2}} \pm\sqrt{\left(\mathrm{2}{x}+\frac{{xy}}{{R}}\right)\left(\mathrm{2}{x}−\frac{{xy}}{{R}}\right)\left(\mathrm{2}{y}+\frac{{xy}}{{R}}\right)\left(\mathrm{2}{y}−\frac{{xy}}{{R}}\right)}}{\mathrm{2}}} \\ $$

Commented by behi83417@gmail.com last updated on 06/Dec/18

wow! perfect sir.thanks in advance.

$${wow}!\:{perfect}\:{sir}.{thanks}\:{in}\:{advance}. \\ $$

Commented by mr W last updated on 07/Dec/18

check with x=y=(√2)R  ⇒a=(√((2(4R^2 )−(2R)^2 ±(√((2(√2)R+2R)^2 (2(√2)R−2R)^2 )))/2))  ⇒a=(√((4R^2 ±4R^2 )/2))=2R or 0⇒correct

$${check}\:{with}\:{x}={y}=\sqrt{\mathrm{2}}{R} \\ $$$$\Rightarrow{a}=\sqrt{\frac{\mathrm{2}\left(\mathrm{4}{R}^{\mathrm{2}} \right)−\left(\mathrm{2}{R}\right)^{\mathrm{2}} \pm\sqrt{\left(\mathrm{2}\sqrt{\mathrm{2}}{R}+\mathrm{2}{R}\right)^{\mathrm{2}} \left(\mathrm{2}\sqrt{\mathrm{2}}{R}−\mathrm{2}{R}\right)^{\mathrm{2}} }}{\mathrm{2}}} \\ $$$$\Rightarrow{a}=\sqrt{\frac{\mathrm{4}{R}^{\mathrm{2}} \pm\mathrm{4}{R}^{\mathrm{2}} }{\mathrm{2}}}=\mathrm{2}{R}\:{or}\:\mathrm{0}\Rightarrow{correct} \\ $$

Answered by behi83417@gmail.com last updated on 06/Dec/18

Commented by behi83417@gmail.com last updated on 06/Dec/18

∡FBC=∡FDC=90^•   BF^2 =4r^2 −x^2 ,DF^2 =4r^2 −y^2   by using petolemy′s teorem in:FDCB:  BD.CF=BC.DF+BF.CD  2r.BD=x.(√(4r^2 −y^2 ))+y.(√(4r^2 −x^2 ))  ⇒BD=((x.(√(4r^2 −y^2 ))+y.(√(4r^2 −x^2 )))/(2r)) .■

$$\measuredangle{FBC}=\measuredangle{FDC}=\mathrm{90}^{\bullet} \\ $$$${BF}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} ,{DF}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$${by}\:{using}\:\boldsymbol{\mathrm{petolemy}}'\boldsymbol{\mathrm{s}}\:{teorem}\:{in}:{FDCB}: \\ $$$${BD}.{CF}={BC}.{DF}+{BF}.{CD} \\ $$$$\mathrm{2}{r}.{BD}={x}.\sqrt{\mathrm{4}{r}^{\mathrm{2}} −{y}^{\mathrm{2}} }+{y}.\sqrt{\mathrm{4}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{BD}=\frac{\boldsymbol{\mathrm{x}}.\sqrt{\mathrm{4}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} }+\boldsymbol{\mathrm{y}}.\sqrt{\mathrm{4}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}{\mathrm{2}\boldsymbol{\mathrm{r}}}\:.\blacksquare \\ $$

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].

$${chech}\:{for}:{x}={y}={r}\sqrt{\mathrm{2}} \\ $$$${BD}=\frac{{r}\sqrt{\mathrm{2}}.{r}\sqrt{\mathrm{2}}+{r}\sqrt{\mathrm{2}}.{r}\sqrt{\mathrm{2}}}{\mathrm{2}{r}}=\frac{\mathrm{4}{r}^{\mathrm{2}} }{\mathrm{2}{r}}=\mathrm{2}{r}\:\left[{ok}\right]. \\ $$

Answered by behi83417@gmail.com last updated on 07/Dec/18

∡BAC=α,∡DAC=β  cosα=((2r^2 −x^2 )/(2r^2 )),sinα=(x/(2r^2 ))(√(4r^2 −x^2 ))  cosβ=((2r^2 −y^2 )/(2r^2 )),sinβ=(y/(2r^2 ))(√(4r^2 −y^2 ))  cos(α+β)=(((2r^2 −x^2 )(2r^2 −y^2 ))/(4r^4 ))−((xy)/(4r^4 ))(√((4r^2 −x^2 )(4r^2 −y^2 )))  BD^2 =2r^2 −2r^2 .cos(α+β)=  =((xy.(√((4r^2 −x^2 )(4r^2 −y^2 )))−(2r^2 −x^2 )(2r^2 −y^2 )+4r^4 )/(2r^2 )).  =((xy.(√((4r^2 −x^2 )(4r^2 −y^2 )))+2r^2 (x^2 +y^2 )−x^2 y^2 )/(2r^2 )).  BD=(√(((xy)/(2r^2 )).(√((4r^2 −x^2 )(4r^2 −y^2 )))+(x^2 +y^2 )−(((xy)/(r(√2))))^2 )).

$$\measuredangle{BAC}=\alpha,\measuredangle{DAC}=\beta \\ $$$${cos}\alpha=\frac{\mathrm{2}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }{\mathrm{2}{r}^{\mathrm{2}} },{sin}\alpha=\frac{{x}}{\mathrm{2}{r}^{\mathrm{2}} }\sqrt{\mathrm{4}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$${cos}\beta=\frac{\mathrm{2}{r}^{\mathrm{2}} −{y}^{\mathrm{2}} }{\mathrm{2}{r}^{\mathrm{2}} },{sin}\beta=\frac{{y}}{\mathrm{2}{r}^{\mathrm{2}} }\sqrt{\mathrm{4}{r}^{\mathrm{2}} −{y}^{\mathrm{2}} } \\ $$$${cos}\left(\alpha+\beta\right)=\frac{\left(\mathrm{2}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)\left(\mathrm{2}{r}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}{\mathrm{4}{r}^{\mathrm{4}} }−\frac{{xy}}{\mathrm{4}{r}^{\mathrm{4}} }\sqrt{\left(\mathrm{4}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)\left(\mathrm{4}{r}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)} \\ $$$${BD}^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} .{cos}\left(\alpha+\beta\right)= \\ $$$$=\frac{\boldsymbol{\mathrm{xy}}.\sqrt{\left(\mathrm{4}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\left(\mathrm{4}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right)}−\left(\mathrm{2}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\left(\mathrm{2}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right)+\mathrm{4}\boldsymbol{\mathrm{r}}^{\mathrm{4}} }{\mathrm{2}\boldsymbol{\mathrm{r}}^{\mathrm{2}} }. \\ $$$$=\frac{\boldsymbol{\mathrm{xy}}.\sqrt{\left(\mathrm{4}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\left(\mathrm{4}\boldsymbol{\mathrm{r}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right)}+\mathrm{2}{r}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{\mathrm{2}\boldsymbol{\mathrm{r}}^{\mathrm{2}} }. \\ $$$${BD}=\sqrt{\frac{{xy}}{\mathrm{2}{r}^{\mathrm{2}} }.\sqrt{\left(\mathrm{4}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)\left(\mathrm{4}{r}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}+\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left(\frac{{xy}}{{r}\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }. \\ $$

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