All lateral edges of the pyramid are equal to $b$, and the height is equal to $h$. Find the radius of the circumscribed circle around the base.

#
Since the lateral edges of the pyramid are equal, its height $S O$ passes through the center $O$ of the circle circumscribed around the base. Let $A$ be one of the vertices of the base of the pyramid. Then $O A$ is the desired radius. From the right triangle $S O A$, we find that

$$
O A=\sqrt{S A^{2}-S O^{2}}=\sqrt{b^{2}-h^{2}}
$$

## Answer

$\sqrt{b^{2}-h^{2}}$