Question Number 199391 by Calculusboy last updated on 03/Nov/23
Answered by AST last updated on 03/Nov/23
$$\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({y}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{10}\Rightarrow\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left(\mathrm{5}−\mathrm{2}{x}\right)^{\mathrm{2}} =\mathrm{10} \\ $$$$\Rightarrow\mathrm{5}\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{8}\right)=\mathrm{0}\Rightarrow{x}=\mathrm{4}\:{or}\:\mathrm{2}\Rightarrow\left({x},{y}\right)=\left(\mathrm{4},\mathrm{2}\right);\left(\mathrm{2},\mathrm{6}\right) \\ $$$$ \\ $$$$\left(\mathrm{5},\mathrm{5}\right);\left(\mathrm{4},\mathrm{2}\right)\Rightarrow{slope}\:{of}\:{tangent}\:{line}\:{passing}\:{through} \\ $$$$\left(\mathrm{4},\mathrm{2}\right)=\frac{−\mathrm{1}}{\mathrm{3}}\Rightarrow\frac{{y}−\mathrm{2}}{{x}−\mathrm{4}}=\frac{−\mathrm{1}}{\mathrm{3}}\Rightarrow\mathrm{3}{y}−\mathrm{6}=−{x}+\mathrm{4}\Rightarrow\mathrm{3}{y}+{x}=\mathrm{10} \\ $$$$\left(\mathrm{5},\mathrm{5}\right);\left(\mathrm{2},\mathrm{6}\right)\Rightarrow{slope}\:{of}\:{tangent}\:{line}\:{passsing}\:\left(\mathrm{2},\mathrm{6}\right) \\ $$$$=\mathrm{3}\Rightarrow\frac{{y}−\mathrm{6}}{{x}−\mathrm{2}}\Rightarrow{y}=\mathrm{3}{x} \\ $$$${x}=\mathrm{1};{y}=\mathrm{3}\:{at}\:{the}\:{point}\:{of}\:{intersection}\:{of}\:{the}\:{two} \\ $$$${tangents} \\ $$
Commented by Calculusboy last updated on 03/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$