Question Number 29332 by Tinkutara last updated on 07/Feb/18
Answered by ajfour last updated on 07/Feb/18
$${let}\:{eq}.\:{of}\:{circle}\:{be} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{gx}+\mathrm{2}{fy}=\mathrm{0}\:\:\:\:\:\left({c}=\mathrm{0}\right) \\ $$$${for}\:{two}\:{circles}\:{to}\:{cut}\:{orthogonally} \\ $$$$\:\:\:\mathrm{2}{g}_{\mathrm{1}} {g}_{\mathrm{2}} +\mathrm{2}{f}_{\mathrm{1}} {f}_{\mathrm{2}} ={c}_{\mathrm{1}} +{c}_{\mathrm{2}} \:\:,\:{so} \\ $$$$\:\:\:\:−\mathrm{4}{g}+\mathrm{6}{f}=\mathrm{10}\:\:\:\:,\:{and} \\ $$$$\:\:\:\:\:\:\mathrm{12}{f}=\mathrm{6}\:\:\:\Rightarrow\:\:\mathrm{2}{f}\:=\mathrm{1} \\ $$$$\:\:\:{and}\:\:\mathrm{2}{g}=−\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${eq}.\:{of}\:{circle}\:{is}\:{therefore} \\ $$$$\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\frac{\mathrm{7}{x}}{\mathrm{2}}+{y}=\mathrm{0}\:\: \\ $$$${or}\:\:\:\:\:\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{7}\boldsymbol{{x}}+\mathrm{2}\boldsymbol{{y}}=\mathrm{0}\:\:. \\ $$
Commented by ajfour last updated on 08/Feb/18
$${is}\:{it}\:{correct},\:{if}\:{so}\:{did}\:{you}\:{follow}\:? \\ $$
Commented by Tinkutara last updated on 08/Feb/18
Yes correct surely Thanks! But if we don't remember this condition we can't do this? Are we to learn all these conditions in Conic Sections?