Question Number 39626 by ajfour last updated on 08/Jul/18
Commented by ajfour last updated on 09/Jul/18
$${Find}\:\boldsymbol{{b}}\:{in}\:{terms}\:{of}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{R}}\:. \\ $$$${What}\:{is}\:\boldsymbol{{a}}\:{in}\:{terms}\:{of}\:\boldsymbol{{R}}\:{if}\:{a}+{b}={R}. \\ $$
Answered by ajfour last updated on 09/Jul/18
$${eq}.\:{of}\:{ABC}\::\:\:\:\:\:−\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:\:\:{y}\:=\:{b}\left(\mathrm{1}+\frac{{x}}{{a}}\right) \\ $$$${eq}.\:{of}\:{circle}:\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${Let}\:{C}\equiv\left({h},{k}\right) \\ $$$${h}^{\mathrm{2}} +{b}^{\mathrm{2}} \left(\mathrm{1}+\frac{{h}}{{a}}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:\:\:\:\:\:….\left({i}\right) \\ $$$${E}\left(\frac{{R}}{\mathrm{2}},\:\frac{{b}}{\mathrm{2}}\right)\:\:{lies}\:{in}\:{line}\:{CEF} \\ $$$${eq}.\:{of}\:{CEF}\: \\ $$$$\:\:\:\:\:\:\:{y}=\left(\frac{\frac{{b}}{\mathrm{2}}+{R}}{{R}/\mathrm{2}}\right){x}−{R} \\ $$$${C}\:{lies}\:{on}\:{both}\:{lines},\:{hence} \\ $$$${b}\left(\mathrm{1}+\frac{{h}}{{a}}\right)=\left(\frac{{b}}{{R}}+\mathrm{2}\right){h}−{R} \\ $$$${h}\left(\frac{{b}}{{R}}−\frac{{b}}{{a}}+\mathrm{2}\right)={b}+{R} \\ $$$${h}=\:\frac{{b}+{R}}{\left(\frac{{b}}{{R}}−\frac{{b}}{{a}}+\mathrm{2}\right)} \\ $$$${let}\:\frac{{b}}{{a}}={z}\:\:\:\:{and}\:\:\frac{{R}}{{a}}={p} \\ $$$${then}\:\:\:\boldsymbol{{h}}\:=\:\frac{\boldsymbol{{a}}\left(\boldsymbol{{z}}+\boldsymbol{{p}}\right)}{\frac{\boldsymbol{{z}}}{\boldsymbol{{p}}}−\boldsymbol{{z}}+\mathrm{2}}\:=\:\frac{\boldsymbol{{a}}\left(\boldsymbol{{pz}}+\boldsymbol{{p}}^{\mathrm{2}} \right)}{\left(\mathrm{1}−\boldsymbol{{p}}\right)\boldsymbol{{z}}+\mathrm{2}\boldsymbol{{p}}} \\ $$$${substituting}\:{in}\:\left({i}\right) \\ $$$$\:\:\left[\frac{{a}\left({pz}+{p}^{\mathrm{2}} \right)}{\left(\mathrm{1}−{p}\right){z}+\mathrm{2}{p}}\right]^{\mathrm{2}} +{b}^{\mathrm{2}} \left[\mathrm{1}+\frac{{pz}+{p}^{\mathrm{2}} }{\left(\mathrm{1}−{p}\right){z}+\mathrm{2}{p}}\right]^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\:\:\left[\frac{\left({pz}+{p}^{\mathrm{2}} \right)}{\left(\mathrm{1}−{p}\right){z}+\mathrm{2}{p}}\right]^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} \left[\mathrm{1}+\frac{{pz}+{p}^{\mathrm{2}} }{\left(\mathrm{1}−{p}\right){z}+\mathrm{2}{p}}\right]^{\mathrm{2}} =\boldsymbol{{p}}^{\mathrm{2}} \\ $$$${solving}\:{above}\:{eq}.\:{we}\:{would}\:{get} \\ $$$$\:\:\:\:\:\:\boldsymbol{{z}}\:=\:\frac{\boldsymbol{{b}}}{\boldsymbol{{a}}}\:{in}\:{terms}\:{of}\:\boldsymbol{{p}}\:=\frac{\boldsymbol{{R}}}{\boldsymbol{{a}}} \\ $$$${but}\:{if}\:{a}+{b}={R}\:\:{then}\:\:{a}+{az}\:=\:{ap} \\ $$$$\Rightarrow\:\:\:\:\:{z}\:=\:{p}−\mathrm{1} \\ $$$$\Rightarrow\:\:\:\frac{{p}^{\mathrm{2}} \left(\mathrm{2}{p}−\mathrm{1}\right)^{\mathrm{2}} }{\left[\mathrm{2}{p}−\left(\mathrm{1}−{p}\right)^{\mathrm{2}} \right]^{\mathrm{2}} }\:+\left({p}−\mathrm{1}\right)^{\mathrm{2}} \:\left[\mathrm{1}+\frac{{p}\left(\mathrm{2}{p}−\mathrm{1}\right)}{\mathrm{2}{p}−\left(\mathrm{1}−{p}\right)^{\mathrm{2}} }\right]^{\mathrm{2}} ={p}^{\mathrm{2}} \\ $$$${p}^{\mathrm{2}} \left(\mathrm{2}{p}−\mathrm{1}\right)^{\mathrm{2}} +\left({p}−\mathrm{1}\right)^{\mathrm{2}} \left({p}^{\mathrm{2}} +\mathrm{3}{p}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={p}^{\mathrm{2}} \left(\mathrm{4}{p}−\mathrm{1}−{p}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${p}^{\mathrm{3}} \left(\mathrm{6}{p}−{p}^{\mathrm{2}} −\mathrm{2}\right)\left(\mathrm{2}−{p}\right)=\left({p}^{\mathrm{3}} +\mathrm{2}{p}^{\mathrm{2}} −\mathrm{4}{p}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{p}^{\mathrm{6}} −\mathrm{8}{p}^{\mathrm{5}} +\mathrm{14}{p}^{\mathrm{4}} −\mathrm{4}{p}^{\mathrm{3}} ={p}^{\mathrm{6}} +\mathrm{4}{p}^{\mathrm{4}} +\mathrm{16}{p}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:+\mathrm{1}+\mathrm{4}{p}^{\mathrm{5}} −\mathrm{16}{p}^{\mathrm{3}} −\mathrm{8}{p}−\mathrm{8}{p}^{\mathrm{4}} +\mathrm{4}{p}^{\mathrm{2}} +\mathrm{2}{p}^{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{12}{p}^{\mathrm{5}} −\mathrm{18}{p}^{\mathrm{4}} −\mathrm{10}{p}^{\mathrm{3}} +\mathrm{20}{p}^{\mathrm{2}} −\mathrm{8}{p}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\:\boldsymbol{{p}}\:\approx\:\mathrm{1}.\mathrm{49438} \\ $$$$\:\:\:\:{and}\:\:\:\frac{\boldsymbol{{a}}}{\boldsymbol{{R}}}=\frac{\mathrm{1}}{\boldsymbol{{p}}}\:\:\Rightarrow\:\:\boldsymbol{{a}}\approx\:\mathrm{0}.\mathrm{669}{R}\:. \\ $$
Commented by ajfour last updated on 09/Jul/18
$${MrW}\mathrm{3}\:{Sir},\:{please}\:{help}\:{check}\:{this} \\ $$$${solution}.. \\ $$
Commented by MrW3 last updated on 10/Jul/18
$${I}\:{couldn}'{t}\:{find}\:{any}\:{mistake}. \\ $$
Commented by ajfour last updated on 10/Jul/18
$${Thank}\:{you}\:{so}\:{much}\:{Sir};\:{it}\:{is} \\ $$$${correct}\:{then}. \\ $$
Answered by MJS last updated on 10/Jul/18
$${R}=\mathrm{1} \\ $$$${A}=\begin{pmatrix}{−{a}}\\{\mathrm{0}}\end{pmatrix}{B}=\begin{pmatrix}{\mathrm{0}}\\{{b}}\end{pmatrix}\:{D}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix}\:{E}=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{{b}}{\mathrm{2}}}\end{pmatrix}\:\:{F}=\begin{pmatrix}{\mathrm{0}}\\{−\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{line}\:{AB} \\ $$$${l}_{\mathrm{1}} :\:{y}=\frac{{b}}{{a}}{x}+{b} \\ $$$$\mathrm{line}\:{EF} \\ $$$${l}_{\mathrm{2}} :\:{y}=\left({b}+\mathrm{2}\right){x}−\mathrm{1} \\ $$$${C}={l}_{\mathrm{1}} \cap{l}_{\mathrm{2}} =\begin{pmatrix}{\frac{{a}\left({b}+\mathrm{1}\right)}{{a}\left({b}+\mathrm{2}\right)−{b}}}\\{\frac{\left({a}\left({b}+\mathrm{2}\right)+\mathrm{1}\right){b}}{{a}\left({b}+\mathrm{2}\right)−{b}}}\end{pmatrix} \\ $$$${C}\:\mathrm{on}\:\mathrm{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{I}\right)\:\:\frac{{a}\left({b}+\mathrm{1}\right)\left({a}\left({b}^{\mathrm{3}} +\mathrm{3}{b}^{\mathrm{2}} +{b}−\mathrm{3}\right)+\mathrm{2}{b}\left({b}+\mathrm{2}\right)\right)}{\left({a}\left({b}+\mathrm{2}\right)−{b}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\:\:\:\:\:\left[{a}=\mathrm{0}\:\vee\:{b}=−\mathrm{1}\:\mathrm{not}\:\mathrm{of}\:\mathrm{further}\:\mathrm{interest}\right] \\ $$$$\:\:\:\:\:\left[{a}\left({b}+\mathrm{2}\right)−{b}\neq\mathrm{0}\:\Rightarrow\:{a}\neq\frac{{b}}{{b}+\mathrm{2}}\:\Leftrightarrow\:{b}\neq\frac{\mathrm{2}{a}}{\mathrm{1}−{a}}\right] \\ $$$$\mathrm{this}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{for}\:{a} \\ $$$${a}=\frac{\mathrm{2}{b}\left({b}+\mathrm{2}\right)}{\mathrm{3}−{b}−\mathrm{3}{b}^{\mathrm{2}} −{b}^{\mathrm{3}} }\:\Rightarrow\:{C}=\begin{pmatrix}{\frac{\mathrm{2}\left({b}+\mathrm{2}\right)}{{b}^{\mathrm{2}} +\mathrm{4}{b}+\mathrm{5}}}\\{\frac{\left({b}+\mathrm{1}\right)\left({b}+\mathrm{3}\right)}{{b}^{\mathrm{2}} +\mathrm{4}{b}+\mathrm{5}}}\end{pmatrix} \\ $$$$\mathrm{but}\:\mathrm{for}\:{b}\:\mathrm{it}\:\mathrm{leads}\:\mathrm{to} \\ $$$${b}^{\mathrm{3}} +\frac{\mathrm{3}{a}+\mathrm{2}}{{a}}{b}^{\mathrm{2}} +\frac{{a}+\mathrm{4}}{{a}}{b}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{which}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{generally}\:\mathrm{solved} \\ $$$$ \\ $$$$\mathrm{with}\:{a}+{b}=\mathrm{1}\:\Rightarrow\:{b}=\mathrm{1}−{a}\:\mathrm{we}\:\mathrm{get} \\ $$$$\left(\mathrm{I}\right)\:\:\frac{{a}\left({a}−\mathrm{2}\right)\left({a}^{\mathrm{4}} −\mathrm{6}{a}^{\mathrm{3}} +\mathrm{8}{a}^{\mathrm{2}} +\mathrm{6}{a}−\mathrm{6}\right)}{\left({a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\:\:\:\:\:\left[{a}=\mathrm{0}\:\vee\:{a}=\mathrm{2}\:\mathrm{not}\:\mathrm{of}\:\mathrm{further}\:\mathrm{interest}\right] \\ $$$$\:\:\:\:\:\left[{a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{1}\neq\mathrm{0}\:\Rightarrow\:{a}\neq\mathrm{2}−\sqrt{\mathrm{3}}\:\wedge\:{a}\neq\mathrm{2}+\sqrt{\mathrm{3}}\right] \\ $$$${a}^{\mathrm{4}} −\mathrm{6}{a}^{\mathrm{3}} +\mathrm{8}{a}^{\mathrm{2}} +\mathrm{6}{a}−\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{again}\:\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{exactly}\:\mathrm{solved}\:\mathrm{but}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{usual}\:\mathrm{difficulties},\:\mathrm{approximately}\:\mathrm{we}\:\mathrm{get} \\ $$$${a}\approx.\mathrm{669174} \\ $$$$\:\:\:\:\:\left[{a}\approx−.\mathrm{895728}\:\mathrm{not}\:\mathrm{of}\:\mathrm{further}\:\mathrm{interest}\right] \\ $$
Commented by ajfour last updated on 10/Jul/18
$${Thank}\:{you}\:{Sir}.\:{I}\:{was}\:{bit}\:{worried}. \\ $$