Question Number 216105 by mr W last updated on 27/Jan/25
Answered by A5T last updated on 27/Jan/25
$$\mathrm{x}^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \Rightarrow\mathrm{x}^{\mathrm{2}} =\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{2}\underset{−} {+}\sqrt{\mathrm{4}+\mathrm{4}}}{\mathrm{2}}=\mathrm{1}\underset{−} {+}\sqrt{\mathrm{2}}>\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$
Commented by mr W last updated on 27/Jan/25
��
Answered by mehdee7396 last updated on 30/Jan/25
$${o}'\left(\alpha+\mathrm{1},\alpha\right)\:\:\&\:{A}\left(\mathrm{0},\mathrm{1}\right)\:\&\:{B}\left(\alpha+\mathrm{1},\alpha\right) \\ $$$$\Rightarrow{M}\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}},\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right) \\ $$$${C}'\::\:\left({x}−\alpha−\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\alpha^{\mathrm{2}} \\ $$$${M}\in{C}' \\ $$$$\Rightarrow\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}−\alpha−\mathrm{1}\right)^{\mathrm{2}} +\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\alpha^{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\sqrt{\mathrm{2}}+\mathrm{1}\:\checkmark \\ $$$$ \\ $$
Commented by mehdee7396 last updated on 28/Jan/25
