Solution:

Here is a suitable procedure: First draw circles centered at $A$ and $B$ with radius $AB$ and join their two points of intersection to create the perpendicular bisector of $AB$. This line contains all points that have the same distance from $A$ and $B$, so the center of a circle through $A$ and $B$ lies on it. Then, similarly construct the perpendicular bisector of $AC$. Because $A$, $B$, and $C$ are known to lie on a circle, the two perpendicular bisectors are not parallel (or else no point could have the same distance from $A$, $B$, and $C$), so they must meet at a point $O$. Draw a circle centered at $O$ with radius $OA$.

Since $O$ lies on both perpendicular bisectors, this circle passes through $B$ and $C$. Finally, any other circle through the same three points would have to have its center on both perpendicular bisectors, and hence at $O$. Its radius must equal $OA$, implying that it coincides with the constructed circle.