Circumference of circle $=10 \pi$
$$
\begin{aligned}
2 \pi r & =10 \pi \text { (let } r \text { be the radius of circle) } \\
r & =5
\end{aligned}
$$
Now, given that 2 diameters of circle lies along
$$
2 x+3 y+1=0 \text { and } 3 x-y-4=0
$$
$\therefore$ Centre is point of intersection of diameters
$$
\begin{gathered}
2 x+3 y+1=0 ...(i)\\
3 x-y-4=0 \Rightarrow y=3 x-4
\end{gathered}
$$
Put $y=3 x-4$ is Eq. (i), we get
$$
\begin{aligned}
2 x+3(3 x-4)+1 & =0 \\
\Rightarrow \quad 2 x+9 x-12+1 & =0 \\
11 x-11=0, x=1, y & =3-4=-1
\end{aligned}
$$
$\therefore$ Center is $(1,-1)$
Now, equation of circle whose centre is $(1,-1)$ and radius is 5
$$
\begin{aligned}
& \text { [Equation of circle-centre }\left(x_1, y_1\right) \text {, radius }=r \\
& \left.\left(x-x_1\right)^2+\left(y-y_1\right)^2=r^2\right] \\
& \Rightarrow \quad(x-1)^2+(y+1)^2=5^2 \\
& \Rightarrow x^2-2 x+1+y^2+2 y+1=25 \\
& \Rightarrow \quad x^2+y^2-2 x+2 y-23=0 \\
&
\end{aligned}
$$