Segments of the lines $2x+3y=1$ and $4x-3y=11$ are diameters of a circle of area $153.94$ square units. Then, the equation of this circle with integer radius is:

Since the given lines are diameters of the circle, both lines pass through the centre of the circle.

So, the centre is the intersection point of

$2x+3y=1$ .....$(1)$

$4x-3y=11$ .....$(2)$

Adding $(1)$ and $(2)$,

$6x=12$

$x=2$

Put $x=2$ in $(1)$:

$2(2)+3y=1$

$4+3y=1$

$3y=-3$

$y=-1$

So, centre of the circle is

$(2,-1)$

Now, area of circle is

$\pi r^2=153.94$

Since $153.94\approx 49\pi$,

$r^2=49$

$r=7$

Equation of circle is

$(x-2)^2+(y+1)^2=7^2$

$(x-2)^2+(y+1)^2=49$

Expanding,

$x^2-4x+4+y^2+2y+1=49$

$x^2+y^2-4x+2y-44=0$