Step 1: Use the key fact about diameters.
Every diameter of a circle passes through its centre, so the two given lines must cross exactly at the centre. We will find that crossing point.
Step 2: Solve the two lines together.
Add $2x+3y=1$ and $4x-3y=11$: the $y$ terms cancel, giving $6x=12$, so $x=2$. Substitute back: $2(2)+3y=1$ gives $3y=-3$, so $y=-1$. The centre is $(2,-1)$.
Step 3: Turn the area into a radius.
The area is $\pi r^2=153.94$. Taking $\pi\approx\frac{22}{7}$, we get $r^2=153.94\times\frac{7}{22}=49$.
Step 4: Read off the radius.
Since $r^2=49$, the integer radius is $r=7$.
Step 5: Write the circle in centre-radius form.
$(x-2)^2+(y+1)^2=49$.
Step 6: Expand to the general form.
$x^2-4x+4+y^2+2y+1=49$, so $x^2+y^2-4x+2y+5-49=0$, that is $x^2+y^2-4x+2y-44=0$.
\[ \boxed{x^2+y^2-4x+2y-44=0} \]