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Question Number 216105 by mr W last updated on 27/Jan/25

Answered by A5T last updated on 27/Jan/25

$${\mathrm{x}}^2=2{\left(\frac{\mathrm{x}+1}{2}\right)}^2\Rightarrow {\mathrm{x}}^2=\frac{{\mathrm{x}}^2+2\mathrm{x}+1}{2}\Rightarrow {\mathrm{x}}^2-2\mathrm{x}-1=0$$$$\Rightarrow \mathrm{x}=\frac{2\underset{-}{+}\sqrt{4+4}}{2}=1\underset{-}{+}\sqrt{2}>0\Rightarrow \mathrm{x}=1+\sqrt{2}$$

Commented by mr W last updated on 27/Jan/25

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Answered by mehdee7396 last updated on 30/Jan/25

$$o^{\prime }(\alpha +1,\alpha )\mathrm{\&}A(0,1)\mathrm{\&}B(\alpha +1,\alpha )$$$$\Rightarrow M(\frac{\alpha +1}{2},\frac{\alpha +1}{2})$$$$C^{\prime }:{(x-\alpha -1)}^2+y^2={\alpha }^2$$$$M\in C^{\prime }$$$$\Rightarrow {(\frac{\alpha +1}{2}-\alpha -1)}^2+{\left(\frac{\alpha +1}{2}\right)}^2={\alpha }^2$$$$\Rightarrow \alpha =\sqrt{2}+1✓$$$$$$

Commented by mehdee7396 last updated on 28/Jan/25

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