Is there any mathematical method {other than sketching the curve from given expression (or equation)} to determine whether a given expession (or equation) forms a closed area or not?
For example- equation of circle form a closed area but how to check it mathematically without sketching its $x$-$y$ curve ?
Edit: This is a half-answer, because it is for parametric curves. Some implicit curves can be parameterized, in which case this applies. I am not aware of a method for determining if a curve is closed given an implicit equation.
Basically, say you have some $t$ that traces out the curve via $(x(t), y(t))$, and say $t$ starts at $0$. If this is a closed curve in the sense you and I are thinking, then after a certain point it will just trace over the original shape. Let's say it returns to its starting point at $\alpha$ for some real number. Then it is closed if $(x(t), y(t)) = (x(u),y(u))$ if and only if $t = u + k\alpha$ for some integer $k$.
Let's use a circle as an example. We have a circle with radius $1$ is traced by the parametric equation with $t$ starting at $0$ $$(\cos(t), \sin(t))$$ Now we have that $t = 2\pi$ is the first time that we have a repeat point, namely, $$(\cos(2\pi), \sin(2\pi)) = (\cos(0), \sin(0))$$ Actually, for every $t$ between $0$ and $2\pi$, we have $$(\cos(t+2\pi), \sin(t+2\pi)) = (\cos(t), \sin(t))$$ This applies for any multiple of $2\pi$, as it just keeps tracing over the original circle. Therefore, we say that this equation is closed because for $t = u + k*2\pi$, where $k$ is an integer, $$(\cos(t), \sin(t)) = (\cos(u), \sin(u))$$ By requiring that there is no repeating point without tracing over the whole shape again, we have a simple closed curve (think of this like no loops where the curve intersects itself), which means there is a discernable area enclosed by the curve.