2. As shown in Figure $1, \odot O$ is tangent to the sides $AB$ and $AD$ of the square $ABCD$ at points $L$ and $K$, respectively, and intersects side $BC$ at points $M$ and $P$. Given that $BM=8$ cm and $MC=17$ cm, the area of $\odot O$ is $\qquad$ square centimeters.
2. $169 \pi$.

As shown in Figure 5, connect the radii $O K, O L, O M$ of $\odot O$, and let the radius be $r$. Draw $M N \perp O L$ at point $N$.
Then $O N=r-8$.
Also, $B C=B M+M C=25$, so
$$
M N=B L=25-r \text {. }
$$

In the right triangle $\triangle O N M$, by the Pythagorean theorem, we have
$$
r^{2}=(25-r)^{2}+(r-8)^{2}
$$
$\Rightarrow r_{1}=13, r_{2}=53$ (not suitable).
Therefore, the area of $\odot O$ is $169 \pi$ square centimeters.