Question Number 30986 by mondodotto@gmail.com last updated on 01/Mar/18
Answered by rahul 19 last updated on 01/Mar/18
$${y}={mx}+{c}\:…\left(\mathrm{1}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} …\left(\mathrm{2}\right)\: \\ $$$${put}\:{the}\:{value}\:{of}\:{y}\:{in}\:{eq}^{{n}} \:\mathrm{2}, \\ $$$${x}^{\mathrm{2}} +\:\left({mx}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)+\mathrm{2}{mcx}+{c}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${D}=\mathrm{0}\:{as}\:{line}\:{touches}\:{the}\:{circle}\:, \\ $$$$\Rightarrow\:{b}^{\mathrm{2}} −\mathrm{4}{ac}=\mathrm{0} \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\mathrm{4}{ac} \\ $$$$\Rightarrow\mathrm{4}{m}^{\mathrm{2}} {c}^{\mathrm{2}} =\mathrm{4}\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:{c}^{\mathrm{2}} ={a}^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right). \\ $$
Commented by mondodotto@gmail.com last updated on 01/Mar/18
$$\mathrm{thanx}\:\mathrm{but}\:\mathrm{what}\:\mathrm{u}\:\mathrm{mean} \\ $$$$\mathrm{D}=\mathrm{0}\:\:\mathrm{D}\:\mathrm{stand}\:\mathrm{for}\:\mathrm{what}? \\ $$
Commented by rahul 19 last updated on 01/Mar/18
$${discriminant}\:{of}\:{a}\:{quadratic}\:\:{eq}^{{n}} . \\ $$