Example 3 Suppose one edge of a tetrahedron is 6, and the other edges are all 5. Then the radius of the circumscribed sphere of this tetrahedron is $\qquad$ [3]
【Analysis】Let
$$
P A=P B=B C=A B=A C=5, P C=6 .
$$

As shown in Figure 6, establish a spatial rectangular coordinate system.
Then $A(0,0,0), B(5,0,0), C\left(\frac{5}{2}, \frac{5 \sqrt{3}}{2}, 0\right)$,
$$
P A=P B .
$$

Let the projection of $P$ on the base $A B C$ be point $P^{\prime}$, and take the midpoint $Q$ of $A B$. Then $Q\left(\frac{5}{2}, 0,0\right)$, and $P^{\prime}$ is on side $C Q$.
$$
\begin{array}{l}
\text { Let } P\left(\frac{5}{2}, y, h\right) . \\
\text { By } P A=5, P C=6 \\
\Rightarrow\left\{\begin{array}{l}
\left(\frac{5}{2}\right)^{2}+y^{2}+h^{2}=25, \\
\left(y-\frac{5 \sqrt{3}}{2}\right)^{2}+h^{2}=36
\end{array}\right. \\
\Rightarrow(y, h)=\left(\frac{\sqrt{3}}{10}, \frac{2 \sqrt{117}}{5}\right) \\
\Rightarrow P\left(\frac{5}{2}, \frac{\sqrt{3}}{10}, \frac{2 \sqrt{117}}{5}\right) .
\end{array}
$$

Let the center of the sphere $O\left(\frac{5}{2}, \frac{5 \sqrt{3}}{6}, z\right)$ be directly above the centroid $E\left(\frac{5}{2}, \frac{5 \sqrt{3}}{6}, 0\right)$ of $\triangle A B C$.
By $O A=O P$
$$
\begin{array}{l}
\Rightarrow\left(\frac{5}{2}\right)^{2}+\left(\frac{5 \sqrt{3}}{6}\right)^{2}+z^{2} \\
\quad=\left(\frac{\sqrt{3}}{10}-\frac{5 \sqrt{3}}{6}\right)^{2}+\left(\frac{2 \sqrt{117}}{5}-z\right)^{2} \\
\Rightarrow z=\frac{15}{\sqrt{117}} \\
\Rightarrow O\left(\frac{5}{2}, \frac{5 \sqrt{3}}{6}, \frac{15}{\sqrt{117}}\right)
\end{array}
$$
$\Rightarrow$ The radius of the circumscribed sphere $r=|O A|=\frac{20 \sqrt{39}}{39}$.