Given point $A$ on the circle $x^2+y^2=R^2$.
From $A$ passes parallel line to the x-axis.
On this parallel line we Assign from point $A$
a Section with Length $2R$ at the Positive direction of the x-axis and we get the point $P(x,y)$.
How can i prove that $p(x,y)$ is a circle with radius $R$ and center $(2R,0)$,
And if the centers of all circles which tangent to the circle $P(x,y)$ and to the line $x=-t$ is a Canonical parabola so what is the parabola equation and $t$ by $R$?
I tried to sketch but it went wrong.
By construction, point $(x,y)$ belongs to the new set if and only if $(x-2R,y)$ lies on the circle $x^2+y^2=R^2$. This yields the equation $(x-2R)^2+y^2=R^2.
Let $C$ be the circle described above. A circle with center $(x,y)$ and radius $r$ is (externally) tangent to $C$ if and only if the distance from $(x,y)$ to $(2R,0)$ is $r+R$. For this circle to be also tangent to vertical line $X=-t$, we need $x+t=r$. Thus, the distance from $(x,y)$ to $(2R,0)$ is $x+t+R$. From $$(x-2R)^2+y^2=(x+t+R)^2$$ we obtain $$2x(3R+t) = y^2+4R^2-(t+R)^2$$ which is an equation of parabola. I leave it for you to determine what combination of parameters makes this parabola "canonical".