Question Number 54481 by ajfour last updated on 04/Feb/19
Commented by ajfour last updated on 04/Feb/19
$${Find}\:{eq}.\:{of}\:{circle}\:{and}\:{coordinates} \\ $$$${of}\:{points}\:{A},{B},{C},{D}\:{in}\:{terms}\:{of}\:\boldsymbol{{c}}. \\ $$
Answered by MJS last updated on 05/Feb/19
$${A}=\begin{pmatrix}{\sqrt{{c}}}\\{\mathrm{0}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{{p}}\\{{p}^{\mathrm{2}} −{c}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{{q}}\\{{q}^{\mathrm{2}} −{c}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{\sqrt{{c}}}\\{{r}}\end{pmatrix} \\ $$$${c}\geqslant\mathrm{0} \\ $$$$\mathrm{normal}\:\mathrm{in}\:{B}: \\ $$$${n}:\:{y}=−\frac{{x}}{\mathrm{2}{p}}−{c}+{p}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${D}\in{n}\wedge\mid{Dn}\mid={r}=−\frac{\sqrt{{c}}}{\mathrm{2}{p}}−{c}+{p}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:{p}^{\mathrm{5}} −\mathrm{2}{cp}^{\mathrm{3}} +\sqrt{{c}}{p}^{\mathrm{2}} +{c}\left({c}−\mathrm{2}\right){p}+\sqrt{{c}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\left({p}−\sqrt{{c}}\right)^{\mathrm{2}} \left({p}^{\mathrm{3}} +\mathrm{2}\sqrt{{c}}{p}^{\mathrm{2}} +{cp}+\sqrt{{c}}\right)=\mathrm{0} \\ $$$${p}^{\mathrm{3}} +\mathrm{2}\sqrt{{c}}{p}^{\mathrm{2}} +{cp}+\sqrt{{c}}=\mathrm{0} \\ $$$${p}={z}−\frac{\mathrm{2}\sqrt{{c}}}{\mathrm{3}} \\ $$$${z}^{\mathrm{3}} −\frac{{c}}{\mathrm{3}}{z}−\frac{\left(\mathrm{2}{c}−\mathrm{27}\right)\sqrt{{c}}}{\mathrm{27}}=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{has}\:\mathrm{got}\:\mathrm{1}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{0}\leqslant{c}<\frac{\mathrm{27}}{\mathrm{4}}, \\ $$$$\mathrm{2real}\:\mathrm{solutions}\:\mathrm{for}\:{c}=\frac{\mathrm{27}}{\mathrm{4}},\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{for}\:{c}>\frac{\mathrm{27}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{can}\:\mathrm{calculate}\:{p}\:\Rightarrow\:{r},\:{q}\:\mathrm{for}\:\mathrm{given}\:{c} \\ $$