Is circular motion possible for an object $A$ with some initial velocity being attracted by a much larger object? 
A particle $A$ is initially given some velocity and is being attracted by a large object $B$ ($m_A\ll m_B$).

*

*Do there exists an initial velocity so that the particle $A$ describes a circular motion and not collide with the object $B$?

*Is there any mathematical proof for it?

 A: Yes ---  but the motion will never look like your drawing. The initial velocity must be tangential to the desired circle and be given by $mv^2/r=GmM/r^2$ .
A: Yes, this is how satellites work. See Orbital Mechanics.
A: Yes it is possible. To see how we must first learn something about uniform circular motion. This means to move in a circle at constant speed. A particle undergoing uniform circular motion has a position $\vec r(t)$ that looks as follows.
$$\vec r(t)=\pmatrix{r\cos\omega t\\r\sin\omega t}$$
Here $\omega=2\pi f$ where $f$ is the frequency. Let's take two derivatives to obtain the acceleration
\begin{align}
\vec v(t)=\frac{d \vec r}{dt}&=\pmatrix{-r\omega\sin\omega t\\\ \ \ r\omega\cos\omega t}\\
\vec a(t)=\frac{d \vec v}{dt}&=\pmatrix{-r\omega^2\cos\omega t\\-r\omega^2\sin\omega t}=-\omega^2\vec r(t)
\end{align}
If we calculate the lengths of the vectors $\vec v,\vec a$ we get $v=\omega\,r$ and $a=\omega^2\, r$. By eliminating $\omega$ we get the equation
$$a=\frac{v^2}r$$
In other words, if something moves in a circle with constant speed it will accelerate with $a=v^2/r$. Or equivalently, if something accelerates with $a=v^2/r$ in the $-r$ direction it will move in a circle with constant speed.
So to get back to your problem: we want the particle to move in a circle so its acceleration must be equal to $a=v^2/r$. We also know the only force acting on the particle is gravity (this is actually not explicitly in your answer, be sure to add every relevant detail in the future because being able to ask good questions is a valuable skill!). Because the only force is gravity we know the particle accelerates with $a=\frac{Gm_B}{r^2}$. So together this gives the condition
$$\frac{v^2}r=\frac{Gm_B}{r^2}$$
which allows you to solve for $v$. Now I have the following questions for you

*

*Solve for $v$. This is the speed necessary to move in a circle

*Is every value of $r$ allowed? For example in the function $\sqrt{r-b}$ only values of $r>b$ are allowed. If every value is allowed this means that at every radius it is possible to move in a circle if you are moving at the right speed.

*Convince yourself that the gravitiational force points in the right direction for circular motion. For example with a drawing.

A: I think you already made it yourself clear. When the small mass is in the depicted circular orbit and you let it make some rotations after which you reverse time then the small mass can never exit the orbit to arrive at A. And if the backward is not possible then also the forward is impossible (assuming that the small mass has no propulsion device).
A: Yes, we need to have initial velocity otherwise it would just do and collide directly with the larger mass along the line joining centers of the masses.
