Let $PQRS$ be a trapezium with $SR=5$ along $x-2y+1=0$ , $PQ=5$ along $2x-y-1=0$ and area $20$. Find the possible equations of circle inscribed in the trapezium.
$\mathbf{My}$ $\mathbf{approach}$
At first I extended $SR$ and $PQ$ to intersect at $A$. Now the inscribed circle in trapezium will be the incentre of the triangle $APS$ thus formed. Using some basic trigonometry I got radius of the inscribed circle as $2$. But now how to get the equation of the circle.
On
(see figure below)
Here is a solution that doesn't use the area constraint, but gets it as a consequence.
Let $(L_1)$ and $(L_2)$ be the initial lines.
Use the formula expressing the distance of a point $P_0=(x_0,y_0)$ to a straight line $(L)$ with equation $ux+vy+w=0$ (https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line) which is
$$d(P_0,L)=\dfrac{|ux_0+vy_0+w|}{\sqrt{u^2+v^2}}$$
Circle centers $P_0=(x_0,y_0)$ must be at the same distance from $(L_1)$ and $(L_2)$, giving :
$$\dfrac{|x_0-2y_0+1|}{\sqrt{1^2+2^2}}=\dfrac{|2x_0-y_0-1|}{\sqrt{2^2+1^2}}$$
Simplifying by $\sqrt{5}$, and dropping subscripts $0$, one obtains the two equations of the angle bissectors:
$$\begin{cases}x-2y+1=+(2x-y-1)\\x-2y+1=-(2x-y-1)\\\end{cases} \iff \begin{cases}y=-x+2 & \ \ (LB_1)\\y=x & \ \ (LB_2)\\\end{cases}$$
Let us treat the case where the center of the circle is on $(LB_2)$ with equation $y=x$.
As the center's coordinates are $(t,t)$, its distance to initial lines $(L_1)$ and $(L_2)$ (i.e., its radius) is:
$$\tag{1}\dfrac{|t-2t+1|}{\sqrt{5}}=\dfrac{|-t+1|}{\sqrt{5}}$$
Thus the equation of the circle is :
$$\tag{2}(x-t)^2+(y-t)^2=\dfrac{(-t+1)^2}{5} \ \iff \ x^2+y^2-2tx-2ty=\dfrac{-9t^2-2t+1}{5}$$
Let $\alpha$ be the angle between $(L_1)$ and $(LB_2)$. A rather simple computation gives $\cos \alpha = 4/5$, confirming your result that the diameter should be 4 and the radius should be 2 [in this case ; on the other case, with $(LB_1)$, the radius is much smaller, as can be seen on the figure].
Thus we have the supplementary constraint that $JK=SR \cos \alpha$ i.e., $5 \cos \alpha$ must be equal to twice the circle's radius, giving, using (1) :
$$4=2\dfrac{|-t+1|}{\sqrt{5}}.$$
As a conclusion,
$$\tag{3}t=1+2\sqrt{5} \ \ \text{and} \ \ t'=1-2\sqrt{5}.$$
It remains to plug these values of $t$ or $t'$ in $(1)$ to obtain the equations of two of the four circles (the big circle in the figure corresponds to the value of $t$ in (3)).
(these circles being symmetrical with respect to $J$).
One can check afterwards that the area of trapezium PQRS is $20$.
It remains to treat the other cases with centers on the other angle bissector $(LB_1)$.
First, here is a simpler way to find the radius of the inscribed circle:
We know that if a quadrilateral $PQRS$ have an inscribed circle, then $\overline{PQ} + \overline{RS} = \overline{RQ} + \overline{PS}$. Therefore, the perimeter of the quadralateral is 20, and the radius is $\frac{2\times 20}{20} = 2$.
For the coordinate of the center, notice that the lines intersects at $(1,1)$, and the angle bisectors are the lines $y = x$ and $x+y = 2$. So you should try to find the points on the two lines so that their distance to the line $x - 2y + 1 = 0$ is 2. There should be 4 solutions in total.
More specifically, consider the line $y = x$, let the point have coordinate $(1+t, 1+t)$, then its distance to $x - 2y + 1 = 0$ is
$$ \frac{|(1+t)-2(1+t) + 1|}{\sqrt 5} = \frac{|t|}{\sqrt 5} $$
so in this case $t = \pm 2\sqrt5$. The other two solutions can be computed similarly.