I am trying to write the equation of torus by keeping in mind the following figure:
Here circle is unit circle whose revolution around z-axis gives the right torus. Let $P=(x,y,z)$ be arbitrary point on torus, $Q=(x,y,0)$ the projection of $P$ into $XY$ plane. Let $C=(x',y',0)$ the center of moving circle. Then
(1) Distance of $P$ from $C$ is $1$ which is expressed by $(x-x')^2+(y-y')^2+z^2=1.$
(2) Distance of $C$ from origin is $2$, so $x'^2+y'^2=4.$
(3) The points $O$, $C$ and $Q$ are colinear. Hence $xy'=yx'. $
From these thre equations, I was unable to obtain a single equation involving $x,y,z$, namely $$(\sqrt{x^2+y^2}-2)^2+z^2=1.$$
Question: How to obtain last equation from (1),(2),(3)? Am I missing something more for it? From (1) and (2) we get $x^2-2xx'+y^2-2yy'+z^2+4=1.$ Now how to remove $x',y'$ from this using (3)? We can remove one, but how both can be removed? So am I missing any more equation?
(Note that the question of equation of torus is appeared previously, but without following answer of that, I was following picture below and trying in above way to obtain equation; so this is not duplicate of previous questions similar to this.)
You introduced two superfluous variables $x'$, $y'$, together with two extra equations, and now you face a horrible elimination problem. There are a lot of pitfalls here, e.g., introducing fallacious solutions, even if you find a trick to eliminate $x'$ and $y'$.
I propose the following procedure: There is this meridian circle $C_{\rm mer}$ living in the "virtual" $(\rho, z)$ right half plane, whereby $\rho:=\sqrt{x^2+y^2}$. Its equation is $$C_{\rm mer}:\quad (\rho-2)^2+z^2=1\ .\tag{1}$$ This circle is rotated around the $z$-axis and in this way produces your torus $T$. A point $(x,y,z)\in{\mathbb R}^3$ has cylindrical coordinates $(\rho,\phi,z)$ given by $$\rho=\sqrt{x^2+y^2},\quad\phi={\rm arg}(x,y),\quad z=z$$ (we don't need to bother about the expression for $\phi$ here). This point lies on $T$ iff the resulting $\rho$ satisfies the equation $(1)$: $$\bigl(\sqrt{x^2+y^2}-2\bigr)^2+z^2=1\ .$$