Question Number 26159 by Tinkutara last updated on 21/Dec/17
$$\mathrm{If}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:\mathrm{2}{xy}\:+\:\mathrm{2}{x}\:+\:\mathrm{2}{y}\:+\:{k}\:=\:\mathrm{0} \\ $$$$\mathrm{represents}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{straight}\:\mathrm{lines}\:\mathrm{then} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{k}. \\ $$
Answered by ajfour last updated on 21/Dec/17
$${let}\:\left({y}−{m}_{\mathrm{1}} {x}−{c}_{\mathrm{1}} \right)\left({y}−{m}_{\mathrm{2}} {x}−{c}_{\mathrm{2}} \right) \\ $$$$\:\:\:\:={y}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{2}{x}+\mathrm{2}{y}+{k}\:=\mathrm{0} \\ $$$$\Rightarrow\:{m}_{\mathrm{1}} {m}_{\mathrm{2}} =\mathrm{1}\:\:\:\:,\:{m}_{\mathrm{1}} +{m}_{\mathrm{2}} =−\mathrm{2}\:, \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:{m}^{\mathrm{2}} +\mathrm{2}{m}+\mathrm{1}=\mathrm{0} \\ $$$${or}\:\:\:\:{m}_{\mathrm{1}} ={m}_{\mathrm{2}} =−\mathrm{1} \\ $$$$\:\:{further}\:\:\:{m}_{\mathrm{1}} {c}_{\mathrm{2}} +{m}_{\mathrm{2}} {c}_{\mathrm{1}} =\mathrm{2} \\ $$$$\Rightarrow\:\:\:\:{c}_{\mathrm{1}} +{c}_{\mathrm{2}} =−\mathrm{2} \\ $$$${and}\:\:{c}_{\mathrm{1}} {c}_{\mathrm{2}} ={k} \\ $$$$\Rightarrow\:\:\:{c}^{\mathrm{2}} +\mathrm{2}{c}+{k}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\left({c}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}−{k} \\ $$$${but}\:{c}\:{has}\:{to}\:{be}\:{real},\:{so} \\ $$$$\:\:\:\:\:\:\mathrm{1}−{k}\:\geqslant\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:{k}\:\leqslant\:\mathrm{1}\:\:\:{or}\:\:\:{k}\:\in\:\left(−\infty,\mathrm{1}\right]\:.\: \\ $$
Commented by ajfour last updated on 21/Dec/17
i think this solution is reliable.
Commented by Tinkutara last updated on 21/Dec/17
Thank you Sir! This is exactly what I did think but answer given is all real values. Is it correct or not? Because k>1 doesn't give any line.