10. (20 points) Given that $\odot P$ passes through point $A(3,1)$, and is tangent to $\odot M: x^{2}+(y-2)^{2}=4$ and the line $l: y=-\frac{3}{4} x-\frac{11}{4}$. Find the radius of $\odot P$.
10. Let the center of the circle be \( P(m, n) \), with radius \( R \). From the problem, we know that point \( P \) is above the line \( l \).
Thus, \( n > -\frac{3}{4} m - \frac{11}{4} \).
Since circle \( \odot P \) is tangent to line \( l \), we have
\[
R = \frac{\left| \frac{3}{4} m + n + \frac{11}{4} \right|}{\sqrt{\left( \frac{3}{4} \right)^2 + 1^2}} = \frac{3 m + 4 n + 11}{5}.
\]

Also, circle \( \odot P \) passes through point \( A(3,1) \), so
\[
(m-3)^2 + (n-1)^2 = R^2.
\]

From the equation of circle \( \odot M \), we know \( M(0,2) \) and the radius of \( \odot M \) is 2.
(1) When \( \odot P \) is internally tangent to \( \odot M \),
\[
\begin{array}{l}
P M = R - 2 \Rightarrow \sqrt{m^2 + (n-2)^2} = R - 2 \\
\Rightarrow m^2 + (n-2)^2 = (R-2)^2.
\end{array}
\]
Subtracting (2) from (3) gives \( 6 m - 2 n - 10 = -4 R \).

Substituting equation (1) into the above equation, we get
\[
n = 1 - 7 m.
\]

Substituting equation (4) into equation (1), we get
\[
R = 3 - 5 m.
\]

Substituting equations (4) and (5) into equation (2), we get
\[
25 m^2 + 24 m = 0 \Rightarrow m = 0 \text{ or } -\frac{24}{25}.
\]

Therefore, \( R = 3 \) or \( 7 \frac{4}{5} \).
(2) When \( \odot P \) is externally tangent to \( \odot M \),
\[
\begin{array}{l}
P M = R + 2 \Rightarrow \sqrt{m^2 + (n-2)^2} = R + 2 \\
\Rightarrow m^2 + (n-2)^2 = (R+2)^2.
\end{array}
\]
Subtracting (2) from (6) gives \( 6 m - 2 n - 10 = 4 R \), and solving with equation (1) gives
\[
m = \frac{13 R + 9}{15}, \quad n = \frac{3 R - 16}{5}.
\]

Substituting these into equation (6), we get
\[
\begin{array}{l}
5 R^2 - 414 R + 1053 = 0 \\
\Rightarrow R = \frac{207 \pm 36 \sqrt{29}}{5}.
\end{array}
\]