I'm note sure if (homework)
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Hint:
I'm assuming that we're given $R, c, d, x_0, v_x, a_x, y_0, v_y, a_y.$
The squared distance between $P = (c,d)$ and $(x,y)$ is given by $ R^2 = (x - c)^2 + (y - d)^2 \tag{1}$ Substitute the definition of $x = x_o + v_x * t + \frac{1}{2} * a_x * t^2 \tag{2} \\ y = y_o + v_y * t + \frac{1}{2} * a_y * t^2$ in $(1) $
You're left with a polynomial in $t.$ Solve for $t,$ this will give you different values of $t = \{t_1, \ldots, t_4\}.$ Substitute each $t_i$ in $(2)$ to get different $(x,y)$ points.
Update # 1:
I'm too lazy to LaTeX the following equation:
2 2 4 3 (0.25 ay + 0.25 ax ) t + (1.0 vy ay + 1.0 vx ax) t 2 2 2 + (vx + 1.0 xo ax - 1.0 ax c - 1.0 ay c + vy + 1.0 yo ay) t 2 2 2 + (-2. vy c + 2. yo vy + 2. xo vx - 2. vx c) t + xo + 2. c - 1. R 2 - 2. xo c - 2. yo = 0
Anyways, that's the degree $4$ polynomial in $t.$ To find a closed-form expressions for the roots $\{t_1 = \cdots, t_2 = \cdots, t_3 = \cdots, t_4 = \cdots\},$ you will need to do a lot of algebra on this polynomial. For example this.