Inverse coordinates

Coordinates of $p$ :

$p(t) = ( 3 + 2 \cos t, 4 + 2 \sin t)$

Define the unit tangential and normal vectors:

$T = (-\sin t, \cos t) \\ N = (\cos t, \sin t)$

Position of origin relative to $p$ :

$v = (0 - p_x, 0 - p_y)$

Express vector $v$ as a weighted sum of $T$ and $N$ :

$v = \alpha \, T + \beta \, N \\ v_x = \alpha \, T_x + \beta \, N_x \\ v_y = \alpha \, T_y + \beta \, N_y$

Solve the system for $\alpha$ and $\beta$ .

$\alpha = \frac{v_y N_x - v_x N_y}{T_y N_x - T_x N_y} \\ \beta = \frac{v_x - \alpha T_x}{N_x}$

Make a scatter plot of the point $(\beta, \alpha)$ as the parameter $t$ varies. It is a circle of radius $5$ centered at $(\beta, \alpha) = (-2,0)$ . I have attached simulation code as well.

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import math

dt = 2.0*math.pi/(10.0**3.0)

t = 0.0

while t <= 2.0*math.pi:

    px = 3.0 + 2.0*math.cos(t)
    py = 4.0 + 2.0*math.sin(t)

    #######

    Tx = -math.sin(t)
    Ty = math.cos(t)

    Nx = math.cos(t)
    Ny = math.sin(t)

    #######

    vx = 0.0 - px
    vy = 0.0 - py

    #######

    alpha = (vy*Nx - vx*Ny)/(Ty*Nx - Tx*Ny)
    beta = (vx - alpha*Tx)/Nx

    print beta,alpha

    t = t + dt

The unit vector along the direction of motion ( the $y'$ axis ) is along $p'(t)$ , the time derivative of $p(t)$ . Hence the unit vector along the $y'$ is $(-\sin t , \cos t)$ . Moreover, the $x'$ axis is along the vector pointing outward from the center of the given circle, so the unit vector of the $x'$ axis is along $(\cos t , \sin t )$ . This defines our moving reference frame. If $r$ is a point coordinate in the global reference frame and $r'$ is the corresponding coordinate in the $x' y'$ frame, then

$r = p(t) + R(t) r'$

where

$R(t) = \begin{bmatrix} \cos t && -\sin t \\ \sin t && \cos t \end{bmatrix}$

The point of interest is the origin $r = 0$ , and therefore, its coordinate in the $x' y'$ frame is

$r'(t) = - R^{-1}(t) p(t) = - R^T(t) p(t)$

But,

$p(t) = \begin{bmatrix} 3 \\ 4 \end{bmatrix} + 2 \begin{bmatrix} \cos t \\ \sin t \end{bmatrix}$

Substituting this and $R(t)$ into the $r'(t)$ equation yields,

$r'(t) = - R^T(t) \begin{bmatrix} 3 \\ 4 \end{bmatrix} - 2 \begin{bmatrix} 1 \\ 0 \end {bmatrix}$

The first term produces a rotating vector (in the clockwise direction, because of the transpose operation, which essentially negates the angle parameter $t$ ) starting at $(-3, -4)$ , while the second term is a fixed offset by $(-2, 0)$

Hence, the sought curve is a circle of radius $5 = \sqrt{ 3^2 + 4^2}$ , offset by the center vector $(-2, 0 )$ .