Let the given two points be $A$ and $B$, the center of the given circle be $O$, the center of the sought circle be $O_{1}$, and the point of tangency of the two circles be $E$. Draw the line $AB$ and mark the point $F$ where the common tangent intersects the extension of $AB$. If we draw any secant $FDC$ through $F$ to the given circle, then

$$
\overline{FE}^{2}=FA \cdot FB=FD \cdot FC
$$

which proves that the points $A, B, C$, and $D$ lie on the circumference of a circle.

Accordingly, we draw a circle through points $A$ and $B$ that intersects the given circle. The intersection point of the common secant of these two circles and $AB$ gives the points of tangency $E$ and $E^{\prime}$ of the tangents drawn from this point to the given circle. The circles passing through $A, B$, and $E$, and respectively through $A, B$, and $E^{\prime}$, give the sought circles. This problem is one case of the Apollonian problem.

Number of solutions: 20.
Let's construct a circle that passes through two given points and is tangent to a given circle.