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 (y−m_1 x−c_1 )(y−m_2 x−c_2 ) =y^2 +x^2 +2xy+2x+2y+k =0 ⇒ m_1 m_2 =1 , m_1 +m_2 =−2 , ⇒ m^2 +2m+1=0 or m_1 =m_2 =−1 further m_1 c_2 +m_2 c_1 =2 ⇒ c_1 +c_2 =−2 and c_1 c_2 =k ⇒ c^2 +2c+k=0 (c+1)^2 =1−k but c has to be real, so 1−k ≥ 0 k ≤ 1 or k ∈ (−∞,1] .](Q26167.png)
$${let}\:\left({y}−{m}_{\mathrm{1}} {x}−{c}_{\mathrm{1}} \right)\left({y}−{m}_{\mathrm{2}} {x}−{c}_{\mathrm{2}} \right) \\ $$$$\:\:\:\:={y}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{2}{x}+\mathrm{2}{y}+{k}\:=\mathrm{0} \\ $$$$\Rightarrow\:{m}_{\mathrm{1}} {m}_{\mathrm{2}} =\mathrm{1}\:\:\:\:,\:{m}_{\mathrm{1}} +{m}_{\mathrm{2}} =−\mathrm{2}\:, \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:{m}^{\mathrm{2}} +\mathrm{2}{m}+\mathrm{1}=\mathrm{0} \\ $$$${or}\:\:\:\:{m}_{\mathrm{1}} ={m}_{\mathrm{2}} =−\mathrm{1} \\ $$$$\:\:{further}\:\:\:{m}_{\mathrm{1}} {c}_{\mathrm{2}} +{m}_{\mathrm{2}} {c}_{\mathrm{1}} =\mathrm{2} \\ $$$$\Rightarrow\:\:\:\:{c}_{\mathrm{1}} +{c}_{\mathrm{2}} =−\mathrm{2} \\ $$$${and}\:\:{c}_{\mathrm{1}} {c}_{\mathrm{2}} ={k} \\ $$$$\Rightarrow\:\:\:{c}^{\mathrm{2}} +\mathrm{2}{c}+{k}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\left({c}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}−{k} \\ $$$${but}\:{c}\:{has}\:{to}\:{be}\:{real},\:{so} \\ $$$$\:\:\:\:\:\:\mathrm{1}−{k}\:\geqslant\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:{k}\:\leqslant\:\mathrm{1}\:\:\:{or}\:\:\:{k}\:\in\:\left(−\infty,\mathrm{1}\right]\:.\: \\ $$
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.