On page 94 of Stewart's "Galois Theory", 4th edition, he leaves as an exercise the the case of the intersection of two circles.
He is proving that the pythagorean closure $\mathbb{Q}^{py}$ of $\mathbb{Q}$ contains all constructible points. Taking $z$ in the intersection $C(z_1, \left|z_2 - z_3\right|) \cap C(z_4, \left|z_5 - z_6\right|)$ of two circles where $z_j \in \mathbb{Q}^{py}$ he arrives at
$(z - z_1) (\overline{z} - \overline{z_1}) = r^2$
$(z - z_4) (\overline{z} - \overline{z_4}) = s^2$
where $r = \left|z_2 - z_3\right|$ and $s = \left|z_5 - z_6\right|$. He says that one can solve in $z$ and $\overline{z}$ and get that $z$ satisfies a quadratic equation with coefficients in $\mathbb{Q}^{py}$. I am having trouble proving that, can someone help?
Sure. The main idea is: intersection two circles is the same than intersecting a circle and a line.
If you substract your two equations, you will get a linear equation in $z$ and $\bar{z}$, so you can write $\bar{z}$ under the form $az+b$, where $a$ and $b$ are constructible (easy to see from the constructibility of the $z_j's$), then you can plug the expression of $\bar{z}$ into your first equation to get a quadratic equation satisfied by $z$, where the coefficients are also constructible.