Question Number 29241 by mondodotto@gmail.com last updated on 05/Feb/18
Answered by ajfour last updated on 05/Feb/18
$${k}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+{y}^{\mathrm{2}} +\mathrm{4}{y}−\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}=\mathrm{0} \\ $$$${coeff}.\:{of}\:{x}^{\mathrm{2}} ={coeff}.\:{of}\:{y}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:{k}−\mathrm{4}\:={k}+\mathrm{1} \\ $$$$\Rightarrow\:\:{k}\rightarrow\infty \\ $$$${then}\:{eq}.\:{of}\:{circle}\:{is} \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{0} \\ $$$${centre}\:\:{is}\:{the}\:{origin},\:{radius}\:=\mathrm{0}\:\:. \\ $$