10.020. A rhombus with diagonals of 12 and 6 cm is inscribed in the intersection of two equal circles. Find the radius of the circle.
Solution.

According to the condition, $ABCD$ is a rhombus, $BD=12$ cm, $AC=6$ cm (Fig. 10.20). Let $O_1O=x, R$ be the radius of the circles. Then $x=R-\frac{AC}{2}=R-3$. In $\triangle O_1OB \quad \angle O_1OB=90^{\circ}, R^2=36+(R-3)^2, R=\frac{45}{6}=\frac{15}{2} \text{ cm}$.

Answer: $7.5 \text{ cm}$.