\(\displaystyle \\
\texttt{We have } \Delta ABC \texttt{ with: } \\ \\
AB = 3 \\
AC = 4 \\
BC = 5 \\ \\
AB^2 + AC^2 = 3^2+4^2 = 9 + 16 = 25 = BC^2 \\
\Longrightarrow~~ \ \textless \ A = 90^o \\
\Longrightarrow~~ BC = \texttt{hypotenuse} \\
\Longrightarrow~~ \texttt{diameter of the circle: } D = BC = 5 \\ \\
\Longrightarrow~~R = \frac{D}{2} = \frac{5}{2} = 2,5 \\ \\
\Longrightarrow~~\texttt{circumference of the circle: } = 2\pi R = \pi D = \boxed{5\pi}\)
