Question Number 138548 by I want to learn more last updated on 14/Apr/21
Answered by nimnim last updated on 14/Apr/21
$${Let}\:{the}\:{centre}\:{of}\:{the}\:{circle}\:{be}\:{C}\left({h},{k}\right) \\ $$$${Since},\:{it}\:{touches}\:{the}\:{x}−{axis}\:{at}\:{P}\left(−\mathrm{3},\mathrm{0}\right), \\ $$$$\:{CP}\bot\:{x}−{axis}\:{and}\:{h}=−\mathrm{3} \\ $$$${since},\:{the}\:{centre}\:{lies}\:{on}\:{the}\:{line}\:\mathrm{2}{x}+{y}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\left(−\mathrm{3}\right)+{k}+\mathrm{1}=\mathrm{0}\Rightarrow{k}=\mathrm{5} \\ $$$$\left({a}\right)\:{coordinate}\:{of}\:{C}=\left(−\mathrm{3},\mathrm{5}\right) \\ $$$$\left({b}\right)\:{radius}=\:\mathrm{5}{units}\:\:\left(\because{CP}\bot{x}−{axis}\right) \\ $$$$\left({c}\right)\:{equation}\:{of}\:{the}\:{circle}: \\ $$$$\:\:\:\:\:\:\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow\left({x}+\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9}+{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{25}=\mathrm{25} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{10}{y}+\mathrm{9}=\mathrm{0} \\ $$$$ \\ $$
Commented by I want to learn more last updated on 14/Apr/21
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}. \\ $$