Question-216105

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

Answered by A5T last updated on 27/Jan/25

x^{2} = 2 {(\frac{x + 1}{2})}^{2} \Rightarrow x^{2} = \frac{x^{2} + 2 x + 1}{2} \Rightarrow x^{2} - 2 x - 1 = 0

\Rightarrow x = \frac{2 \underline{+} \sqrt{4 + 4}}{2} = 1 \underline{+} \sqrt{2} > 0 \Rightarrow 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^{'} (α + 1, α) & A (0, 1) & B (α + 1, α)

\Rightarrow M (\frac{α + 1}{2}, \frac{α + 1}{2})

C^{'} : {(x - α - 1)}^{2} + y^{2} = α^{2}

M \in C^{'}

\Rightarrow {(\frac{α + 1}{2} - α - 1)}^{2} + {(\frac{α + 1}{2})}^{2} = α^{2}

\Rightarrow α = \sqrt{2} + 1 ✓

Commented by mehdee7396 last updated on 28/Jan/25