On
As comments to your question discuss various alternatives, here is the way I read your problem: you have two infinite lines meeting at a right angle. Without loss of generality let's assume them to be the two coordinate axes. And you have a line of fixed length, whose one end point is on one of the lines and whose other end point is on the other of the lines. Let's assume that the length of this line is $2$, so that you get a circle of radius $1$ constructed over that line. Never mind that the circle itself intersects the two orthogonal lines; it's only the end points of the diameter which slide, not the circle itself.
Thales's theorem tells you that if you look at the triangle formed by the end points of the diameter and the point where the fixed lines intersect, that triangle is a right-angled triangle and therefore the point where the fixed lines intersect has to lie on the circle as well. So all the positions of the sliding circle can be characterized by the point of intersection (the origin in my above choice of lines) being a point on the circle.
What is the envelope of a circle of radius $1$ rotating around a point on its boundary? It's a circle of radius $2$. In the more general setup, you get a circle whose radius is equal to the diameter that slides, centered on the point where the two fixed lines intersect.
Things would be more complicated if you had rays instead of lines, i.e. if you were to restrict the sliding endpoints to not move beyond the point of intersection. But I see nothing in your problem statement to suggest that extra complexity.
I understood the question as:
Find the envelope of the semicircle whose diameter is a ladder of length $2R$ sliding between a wall and floor.
A steel semicircle is welded to its diameter which is the sliding ladder.
Inclination to floor/ horizontal is $\theta$. The locus of the ladder mid-point during sliding is a quarter of a circle.. which can be separately proved if needed.
$$ ( x- R\cos \theta)^2 + (y-R \sin \theta)^2=R^2 \tag1 $$
Partially differentiate with respect to $\theta$ the parameter or angle which the ladder makes to floor when sliding. The C-discriminant method is used:
$$2(x-R \cos\theta) \cdot R\sin\theta+ 2 (y-R\sin \theta) \cdot-R \cos\theta =0 \tag2$$
Simplifying
$$ \tan \theta =\frac{y}{x}\tag 3 $$
The above is suggestive of a circle envelope but is not yet conclusive..
Plug the above into 1) to eliminate $ \theta. $ The sliding semicircle envelope equation after simplification becomes:
$$ (x^2+y^2) \left( 1-\frac{R}{\sqrt{x^2+y^2}}\right)^2 =R^2 \tag 4 $$
There is another envelope for the bottom side semicircle, which physically interferes with the wall..so not given now but can be included for a more complete answer.
A geometrical construction $R=1$ gives a sense that the envelope is a quarter of a circle. Still checking my work. Hope no errors in understanding the question, formulation or typing.
A rough sketched envelope: