2. As shown in the figure, $\odot O$ is tangent to the sides $A B, A D$ of the square $A B C D$ at points $L, K$, respectively, and intersects side $B C$ at points $M, P$, with $B M=8$ cm, $M C=17$ cm. Then the area of $\odot O$ is $\qquad$ square cm.
Answer
As shown in the figure, the side length of square $ABCD$ is $25, BM=8$. Let the radius of $\odot O$ be $r$, then $OF=25-r, MF=r-8$, thus $(25-r)^{2}+(r-8)^{2}=r^{2} \Rightarrow r^{2}-66 r+689=0$ $\Rightarrow(r-13)(r-53)=0 \Rightarrow r=13$.
Therefore, the area of $\odot O$ is $169 \pi$ square centimeters.