If the lines $2x+3y+1=0$ and $3x-y-4=0$ lie along diameters of a circle of circumference $10\pi ,$ then the equation of the circle is

The intersection of any two diameters of a circle is its center.
Solving the system of equations for the diameters:$2x+3y=-1$$3x-y=4\Rightarrow y=3x-4$ Substitute $y$ in the first equation:$2x+3(3x-4)=-1$$2x+9x-12=-1\Rightarrow 11x=11\Rightarrow x=1$.
Then $y=3(1)-4=-1$.
The center is $(1,-1)$.The circumference is $2\pi r=10\pi ,$ so the radius $r=5$.
The equation of the circle is $(x-1{)}^2+(y+1{)}^2=5^2$:
$x^2-2x+1+y^2+2y+1=25$
$x^2+y^2-2x+2y-23=0$.