Existence of uniform circular motion 
We know that if any body has some mass and is moving with a velocity v at an instant $t = 0$ , then it is not going to have same velocity after some time if any net force acts on it. It is because the force causes acceleration and acceleration causes change in velocity.

Consider a ball of mass $M$ describing a uniform circular motion in a circle of radius $R$ . This ball is having constant speed. The particle at any instant is having only force (as well as acceleration) towards the centre.
Suppose it had a velocity of magnitude $x$ at the instant it starts the uniform circular motion. From this instant only, the centripetal force works because before this (when it was not doing circular motion) no centripetal force was there.
Now, magnitude of velocity towards the centre is increasing as time is passing because there is a positive acceleration towards the centre. Let it be denoted as $v(t)$ .
The magnitude of resultant velocity is given as :
$$R = \sqrt{(x)^2 + { v(t) } ^2  }$$
as from the Pythagoras theorem.
As $v(t)$ is increasing, thus the resultant's magnitude is also. So, it's not uniform circular motion?
Where did I go wrong ?
I would be glad if the actual description of circular motion is provided. I cannot get any intuitive description about circular motions and it feels very hard.
 A: Let $\mathbf{\hat r}$ be a unit vector from the ball to the centre of the circle. If the ball's velocity is $\mathbf{v}$ then the magnitude of component of component of $\mathbf{v}$ directed towards the centre of the circle is
$$v_r=\mathbf{v}.\mathbf{\hat r}$$
so when we differentiate with respect to time we have
$$\displaystyle \frac {d(v_r)}{dt} = \frac {d \mathbf{v}}{dt} . \mathbf {\hat r} + \mathbf v . \frac {d \mathbf{\hat r}}{dt}$$
I think you are assuming that
$$\displaystyle \frac {d(v_r)}{dt} = \frac {d \mathbf{v}}{dt} . \mathbf {\hat r}$$
but this would only be true if the second term were zero, which requires $\mathbf {\hat r}$ to be a constant vector. But $\mathbf {\hat r}$ is not a constant vector because it is changing direction. In fact, for uniform circular motion, the second term exactly cancels the first and we have $\frac {d (v_r)}{dt}=0$, so $v_r$ is constant - and, furthermore, this constant is zero.
A: 
I would be glad if the actual description of circular motion is provided. I cannot get any intuitive description about circular motions and it feels very hard.

Here I give mathematical description of (non)uniform circular motion in which radius of motion does not change. I will first derive expressions for radial (centripetal) and tangential acceleration components, and then I will address your questions.

Radial and tangential acceleration components
The circular motion path (trajectory) in Cartesian coordinate system is
$$\vec r = x \hat \imath + y \hat \jmath = r_0 \hat r$$
$$x^2 + y^2 = r_0^2, \qquad \boxed{\vec r \cdot \vec r = r_0^2} \tag 1$$
where position vector magnitude (radius) $r_0$ is constant, and $\hat r$ is position unit vector. Now take time derivative of Eq. (1) by noting that $r_0$ is constant
$$x \dot x + y \dot y = 0, \qquad x v_x + y v_y = 0$$
$$\boxed{\vec r \cdot \vec v = 0}, \qquad r_0 v_0 \cos\phi = 0 \tag 2$$
where $\phi$ is angle between position and velocity vectors, and $v_0$ is magnitude of the velocity vector. When radius does not change, the two vectors are always perpendicular, i.e. $\phi = 90^\circ$ or $\phi = -90^\circ$.
Take time derivative of Eq. (2)
$$\dot x \dot x + x \ddot x + \dot y \dot y + y \ddot y = 0, \qquad x a_x + y a_y = -v_0^2$$
$$\boxed{\vec r \cdot \vec a = - \vec v \cdot \vec v}, \qquad r_0 a_0 \cos\rho = -v_0^2 \tag 3$$
where $\rho$ is angle between position and acceleration vectors, and $a_0$ is magnitude of the acceleration vector. Since all magnitudes are positive, the angle is in range $90^\circ \leq \rho \leq 270^\circ$.
Magnitude of the velocity vector is
$$v_x^2 + v_y^2 = v_0^2, \qquad \boxed{\vec v \cdot \vec v = v_0^2} \tag 4$$
Take time derivative of Eq. (4)
$$v_x \dot v_x + v_y \dot v_y = v_0 \dot v_0, \qquad v_x a_x + v_y a_y = v_0 \dot v_0$$
$$\boxed{\vec v \cdot \vec a = v_0 \dot v_0}, \qquad a_0 \cos \alpha = \dot v_0 \tag 5$$
where $\alpha$ is angle between velocity and acceleration vectors. When velocity vector magnitude does not change (uniform circular motion), then $\dot v_0 = 0$ and $\alpha = 90^\circ$ or $\alpha = -90^\circ$ and $\rho = 180^\circ$. This means that net acceleration points towards the center of rotation, i.e. in the direction of $-\hat r$, hence the name radial (centripetal) acceleration. However, if velocity vector magnitude changes (non-uniform circular motion), i.e. $\dot v_0 \neq 0$, then velocity and acceleration vectors are no longer perpendicular and there is acceleration component other than the radial component.
Let's now write velocity and acceleration vectors in terms of tangential and radial components
$$\vec v = v_0 \hat r_\perp, \qquad \vec a = -a_\perp \hat r + a_\parallel \hat r_\perp \tag 6$$
where $a_\perp$ is radial acceleration component that points towards the center of rotation ($-\hat r$), and $a_\parallel$ is tangential acceleration component that is perpendicular to the radial component. Note that $\hat r_\perp$ points in the direction of $\vec v$, and the acceleration vector magnitude is $a_0^2 = a_\perp^2 + a_\parallel^2$. By combining Eqs. (3) and (5) with Eq. (6) we get
$$\vec r \cdot \vec a = -v_0^2, \quad \text{and} \quad \vec r \cdot \vec a = (r_0 \hat r) \cdot (-a_\perp \hat r + a_\parallel \hat r_\perp) = -r_0 a_\perp$$
$$\vec v \cdot \vec a = v_0 \dot v_0 \quad \text{and} \quad \vec v \cdot \vec a = (v_0 \hat r_\perp) \cdot (-a_\perp \hat r + a_\parallel \hat r_\perp) = v_0 a_\parallel$$
where $\hat r \cdot \hat r = 1$ and $\hat r \cdot \hat r_\perp = 0$. From this it follows
$$\boxed{a_\perp = v_0^2/r_0}, \qquad \boxed{a_\parallel = \dot v_0}$$
Note that $a_\perp$ is always positive, i.e. radial component $a_\perp$ always points in the direction of $-\hat r$, whereas sign of $a_\parallel$ depends on the sign of $\dot v_0$. When velocity vector magnitude is increasing, then $\dot v_0 > 0$ and tangential acceleration component points in the same direction as $\vec v$ and vice-versa.

Q&A

Suppose it had a velocity of magnitude x at the instant it starts the uniform circular motion. From this instant only, the centripetal force works because before this (when it was not doing circular motion) no centripetal force was there.

Correct. A force in the direction of the center of rotation (centripetal or radial force) is needed to change direction of motion.

Now, magnitude of velocity towards the centre is increasing as time is passing because there is a positive acceleration towards the centre. Let it be denoted as $v(t)$.

This is not entirely correct. If the force towards the center is exactly equal to $F = v_0^2/r_0$ then the object is only going to change direction, i.e. it will not "fall" towards the center. If the force is larger (stronger) than that, to keep the same velocity $v_0$ the radius $r_0$ will decrease and the object will keep rotating at a smaller radius, and vice-versa for smaller (weaker) forces.

So, it's not uniform circular motion?

If the radial force $F_r$ is not equal to $v_0^2/r_0$, then the radius of circular motion will change and it will eventually settle to new radius $r_0'$ for which the expression $v_0^2/r_0'$ will be equal to the (initial) radial force component $F_r$.
A: You are using the term a*dt for velocity towards center{v(t)} but in that term we assume a=constant in dt time which is not the case in uniform circular motion and changes direction in dt time too.So the net adt in this case changes the direction of velocity
