Here is a construction. There are actually two solutions.
The construction relies on the notion of power of a point with respect to a circle. It can be proved that, for any point $K$ in the plane and any circle with centre $O$ and radius $R$, if a line through $K$ intersects the circle in two points $A$ and $B$, the product $KA\cdot KB$ is constant (and is called the power of $K$ w.r.t. the circle):
This power is $\enspace KA \cdot KB =KT^2 =KO^2- R^2.$
In particular if the line $(KT)\,$ is tangent to the circle at $T$, $KT$ is the geometric means of $KA$ and $KB$.
Thus in order to solve the problem, we have to take the intersection $K$ of the line $(AB)$ with the given line, and construct geometrically the geometric means of $KA$ and $KB$.
This can be done with the right triangle altitude theorem: let $C$ be the symmetric of $A$ ww.r.t. $K$ and $P$ be an intersection of the line perpendicular to $(AB)$ through $K$ and the circle with diameter $[AB]$:
By the right triangle altitudet theorem, $KP^2=KC\cdot KB=KA\cdot KB$.
There remains to take the intersections $T$ and $U$ of the circle with centre $K$ and radius $KP$ with the given line. These will be the points of contact of the circles with the line.
To finish the construction, the centres of the circles are the points of intersection of the lines perpendicular to the given line through $T$ and $U$, with the perpendicular bisector of $[AB]$: