= Solution
A geometric circle is an <unknot>. Compactify $\mathbb R^3$ to $S^3$, and carry the circle to $\{(z_1,z_2)\in\mathbb C^2:|z_1|^2+|z_2|^2=1,\ z_2=0\}$. Its complement has coordinates $z_1\in\Delta$ and $z_2=\sqrt{1-|z_1|^2}e^{i\theta}$, making it homeomorphic to $\Delta\times S^1$. It deformation retracts to $S^1$. Removing the additional point at infinity does not change the <fundamental group>: van Kampen for a small ball about that point uses a <simply connected> punctured ball, homotopy equivalent to $S^2$. Thus
$$
\boxed{\pi_1(\mathbb R^3\setminus A)=\mathbb Z.}
$$
A meridian linking the circle once generates it. This uses “circle” in its geometric sense; the complement of an arbitrary knotted embedded copy of $S^1$ need not have this <group>.
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