How can we find velocity, acceleration etc, of a revolving particle with respect to an observer inside the circle (not at center)? A particle is revolving in horizontal a circle of radius $R$ with constant speed of $|\vec{v}|$ and constant angular velocity $\omega$. There is another observer standing inside the circle, at a distance $x$ from the center of the circle.
Will this observer see the particle revolving around the circle with same velocity, acceleration and angular velocity or will it differ, if it differs then how can we find the function of velocity, acceleration, speed, angular velocity and $\phi$(refer diagram), in time; for this observer in his frame; i.e. how can we find $\vec{v}(t)_\text{observer}\text{, }\vec{a}(t)_\text{observer}\text{, }|\vec{v}|(t)_\text{observer}\text{(speed), }\theta(t)_\text{observer}\; \& \;\phi(t)_\text{observer}$
Assume that an ideal(massless & inextensible) string is attached to the revolving particle and it has mass $m$. At $t=0$ the particle is at a point where the ray from center to observer cuts the circle: initial position(refer diagram) and revolves in  anticlockwise direction.

$$\alpha  = x$$
Edit: can we find a relation between $dx \; \& \;d\phi$
 A: Lets put the observer at A and the particle at B. The position kinematics are:


*

*$\vec{r}_A(t) = (a,0,0)$

*$\vec{r}_B(t) = (r \cos \phi,r \sin \phi,0)$


The velocity kinematics are


*

*$\vec{v}_A(t) = (0,0)$

*$\vec{v}_B(t) = (-r \omega \sin \phi,r \omega \cos \phi)$


The acceleration kinematics are


*

*$\vec{a}_A(t) = (0,0)$

*$\vec{a}_B(t) = (-r \omega^2 \cos \phi,-r \omega^2 \sin \phi)$


provided that $\dot{\omega}=0$.
Now the relative position of B to A is


*

*$\vec{d} = \vec{r}_B - \vec{r}_A = (r \cos\phi-a,r \sin \phi)$

*The polar distance is $d = \|\vec{r}\| = \sqrt{r^2+a^2-2 r a \cos\phi} $

*The polar angle is $\theta = \tan^{-1} \left( \frac{r \sin\phi}{r \cos\phi-a} \right)$


The derivative of the relative position is


*

*The radial speed is $v_\parallel=\dot{d} = \frac{a \omega r \sin\phi}{d}$

*The proper motion is $v_\perp=\sqrt{(\omega r)^2-\dot{d}^2} = \frac{\omega r (a \cos\phi-r)}{d}$


You can differential once more to get the acceleration components.
