9. As shown in Figure 2, in $\triangle A B C$, $A B=B C=5, A C$ $=7, \triangle A B C$ has an inscribed circle $\odot O$ that is tangent to side $A C$ at point $M$. A line $M N$ parallel to side $B C$ is drawn through point $M$ and intersects $\odot O$ at point $N$. A tangent to $\odot O$ is drawn through point $N$ and intersects $A C$ at point $P$. Then $M N-N P=$ $\qquad$ .
9.0.6.

As shown in Figure 4, connect $P O$ intersecting $B C$ at point $D$. By the tangent property, we know
$$
P O \perp M N \text {. }
$$

Since $M N \parallel B C$, it follows that $P O \perp B C$.
Therefore, point $D$ is the point of tangency between $\odot O$ and $B C$.
Also, $C D=C M=3.5, P C=\frac{C D}{\cos C}=3.5 \times \frac{10}{7}=5$,
so $P N=P M=P C-C M=1.5$,
$$
M N=2 P M \cos C=2 \times 1.5 \times \frac{7}{10}=2.1 \text {. }
$$

Thus, $M N-N P=0.6$.