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Consider the circle with radius $2\sqrt{2}$ centered at the origin and the line joining the points $\left(\frac{1}{2}, 0\right)$, $(4\sqrt{2}, \sqrt{2})$. The points of intersection of the line and circle have coordinates which lie in a field extension of $\mathbb{Q}$. Give a precise description of this field extension.

user326752
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    First you must find the points of intersection! – user310540 Apr 05 '16 at 00:39
  • The solutions look horrible: WA. – lhf Apr 05 '16 at 01:35
  • So you solve for the points of intersection and find they are products/sums of rationals and $\sqrt2$ and $\sqrt{121-16\sqrt2}$. So what field extension is needed to accommodate these two values? – almagest Apr 05 '16 at 10:01
  • How do you find out the intersection point. I calculate x = $\frac{4\pm \sqrt{136982-32992\sqrt{2}}}{129-16\sqrt{12}}$ and I'm stuck. – user326752 Apr 07 '16 at 02:15

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