4. Place two circular paper pieces with a radius of 5 and one circular paper piece with a radius of 8 on the table, so that they are pairwise externally tangent. If a large circular paper piece is used to completely cover these three circular paper pieces, then the minimum radius of this large circular paper piece equals $\qquad$ $ـ$.

Put the above text in English, please keep the original text's line breaks and format, and output the translation result directly.
4. As shown in the figure, let the radius of $\odot O_{1}$ be $8$; the radii of $\odot O_{2}$ and $\odot O_{3}$ be $5$, with the tangent point at $A$. By symmetry, the center $O$ of the smallest circular paper piece that can cover these three circles must lie on the line of symmetry $O_{1} A$, and it must be internally tangent to the three given circles. If we set the radius of this circular paper piece to be $r$, then in the right triangle $\triangle O_{1} O_{2} A$, $O_{1} A=\sqrt{(8+5)^{2}-5^{2}}=12$. In the right triangle $\triangle O O_{2} A$, $(r-5)^{2}=5^{2}+(12+8-r)^{2}$, solving for $r$ gives $r=\frac{40}{3}=13 \frac{1}{3}$.