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Home » \( F_12 = 144 \) - Crosslake

\( F_12 = 144 \) - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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Solution:** Résolvez la deuxième équation pour \( x \):
x = y + 2
Substituez dans la première équation :
2(y + 2) + 3y = 6
2y + 4 + 3y = 6
5y + 4 = 6
5y = 2
y = \frac{2}{5}
Substituez de nouveau pour trouver \( x \):
x = \frac{2}{5} + 2 = \frac{2}{5} + \frac{10}{5} = \frac{12}{5}

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This is the equation of a **circle** in the \( (r,z) \)-plane, rotated...
Wait: \(\left(r - rac{c}{2}
ight)^2\) is a circle in the \(r\)-axis plane, centered at \(r = c/2,...
Indeed, expand:
r^2 - c r + rac{c^2}{4} + z^2 = rac{c^2}{4} \Rightarrow r^2 + z^2 =...
But \(r^2 + z^2 = x^2 + y^2 + z^2\), and \(r = \sqrt{x^2 + y^2}\), so:
Now square both sides carefully:
This is not a sphere â it's a **quartic surface**. But wait: earlier...
Let \(r = \sqrt{x^2 + y^2}\), so:
r^2 + z^2 = c r \Rightarrow r^2 - c r + z^2 = 0

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