10.225. Two circles of different radii touch each other externally. Find the angle determined by the chords connecting the point of contact of the circles with the points of contact of their common external tangent.
Solution.

Let $AB$ be the common external tangent of the given circles, $C$ be the point of tangency, and $D$ be the intersection point of $AB$ and the common tangent of the circles at point $C$ (Fig. 10.34). Then $DB = DC$, $DA = DC$, hence $\angle DCB = \angle DBC$, $\angle DAC = \angle DCA$. From $\triangle ACB: \angle DAC + \angle DBC + \angle ACB = 180^{\circ}$, therefore, $\angle DAC + \angle DBC = \angle ACB$ and $2 \angle ACB = 180^{\circ}$, from which $\angle ACB = 90^{\circ}$.

Answer: $90^{\circ}$.