Assume you a circle centred at the origin with radius $r$.
Let $(r+a,0), a>0$ be a point outside the circle.
Let a particle be moving along the circle with angular velocity $\omega$.
I want to find an equation of motion of the external point in the coordinate system of the moving particle?
Assume you have two coordinate frames $\mathcal{K}_0$ and $\mathcal{K}_1$. The first coordinate frame $\mathcal{K}_0$ is fixed at the origin, which corresponds to the centre of the circle. The second coordinate is rotated with the angle $\alpha=\omega t$ and then translated radially such that the origin of this coordinate system is attached to the circle.
I will first consider the homogeneous transformation from the rotating frame to the fixed frame and then invert the given transformation.
$$^{0}T_1=\begin{bmatrix} &^{0}R_1 & & ^{0}d_1\\ 0 & 0 & 0 & 1\\ \end{bmatrix}$$ which is composed by the following two components $$^{0}R_1=\begin{bmatrix} \cos \alpha & -\sin\alpha & 0 \\ \sin \alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$
$$^{0}d_1=\begin{bmatrix} r\cos \alpha\\ r\sin \alpha \\ 0 \\ \end{bmatrix}.$$
The $^{0}R_1$ is representing a rotation about the $z$-axis with an angle of $\alpha$. The $^{0}d_1$ part of this matrix is representing the radial displacement of our coordinate frame.
We know the position of the point $^{0}p=[r+a,0,0]^T$ in the fixed frame and we want to know its coordinates in the coordinate frame on the circle $^{1}p$. We can find this by using
$$^{0}p=^{0}T_1\,^{1}p$$
and inverting $^{0}T_1$, which can be done by using this formula
$$\left[^{0}T_1\right]^{-1}={}^{1}T_0=\begin{bmatrix} &^{0}R^T_1 & & -^{0}R^T_1\,^{0}d_1\\ 0 & 0 & 0 & 1\\ \end{bmatrix}$$
This will give you
$${}^{1}T_0=\begin{bmatrix} \cos \alpha & \sin \alpha & 0 & -r\\ -\sin \alpha & \cos \alpha &0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix}.$$
Hence, we can find the coordinates of the point $p$ in the moving frame by using
$${}^1p = ^{1}T_0{}^0p=[(a+r)\cos \alpha-r,-(a+r)\sin \alpha,0,1]^T $$
Note, that the last coordinate is just added such that we can work with homogenous transformations. And the $z$-coordinate is trivially zero.