2. As shown in Figure 1, given three equal circles $\odot O_{1}$, $\odot O_{2}$, and $\odot O_{3}$ with a common point $H$, points $A$, $B$, and $C$ are the other intersection points of these circles. Then $H$ is the ( ) of $\triangle A B C$.
(A) Circumcenter
(B) Incenter
(C) Orthocenter
(D) Centroid
2.C.

Connect $A H, B H, C H$. Since $\odot \mathrm{O}_{1}, \odot \mathrm{O}_{2},$
$\odot O_{3}$ are three equal circles, then
$$
\begin{array}{l}
\angle A B H=\angle A C H, \angle B A H=\angle B C H, \\
\angle C B H=\angle C A H . \\
\text { Therefore, } \angle B A H+\angle A B C=\angle B A H+\angle A B H+\angle C B H \\
=\frac{1}{2}(\angle B A H+\angle B C H+\angle A B H+ \\
\angle A C H+\angle C B H+\angle C A H) \\
=\frac{1}{2}(\angle B A C+\angle C B A+\angle A C B)=90^{\circ} .
\end{array}
$$

Thus, $A H \perp B C$.
Similarly, $B H \perp A C, C H \perp A B$.
Therefore, $H$ is the orthocenter of $\triangle A B C$.