The following is adapted from an answer I gave to a similar but not identical question.
Let the position of the (unmoving) sun be $(S_x,S_y)$. Drag this position to the origin. Let the current position of the planet be $(A_x,A_y)$. Our dragging procedure drags $(A_x,A_y)$ to $(A_x-S_x,A_y-S_y)$.
For simplicity of notation, let $u=A_x-S_x$, $v=A_y-S_y$. Now we determine the angle that the positive $x$-axis has to be rotated through (counterclockwise) to hit the line from the origin to $(u,v)$. Call this angle $\theta$.
Then $\theta$ is the angle, say in the interval $(-\pi,\pi]$, whose cosine is $u/r$, and whose sine is $v/r$, where $r$ is the radius of the circular orbit. So from now on we can take $\theta$ as a known angle. It can be calculated once and for all from knowledge of $(S_x,S_y)$ and $(A_x,A_y)$. (Be careful to take the signs of $u$ and $v$ into account when calculating $\theta$.)
Assume that we are travelling counterclockwise around the sun, at an angular speed of $\alpha$ radians per time unit. Then after elapsed time $t$, we have travelled through an angle $\alpha t$. (If we are travelling clockwise, use $-\alpha t$.)
After the travel, our angle with the positive $x$-axis is $\theta+\alpha t$. This means that we are at the point with coordinates
$$(r\cos(\theta+\alpha t), \: \: r\sin(\theta+\alpha t)).$$
Now transform back, by adding $(S_x,S_y)$ to the point. We obtain
$$(S_x+r\cos(\theta+\alpha t),\:\: S_y+r\sin(\theta+\alpha t)).$$
This is the position of the planet $t$ time units after it was in position $(A_x,A_y)$.
All the components of this formula are known, so now we can compute.
Comment: By using the addition laws for cosine and for sine, we can obtain
the following alternate (and for many purposes more useful) version of the answer.
Let $(A_x(t), A_y(t)$ be the position of the planet at time $t$. So $(A_x, A_y)$ could also be written as $(A_x(0), A_y(0))$.
$$A_x(t)=S_x+(A_x(0)-S_x)\cos(\alpha t)-(A_y(0)-S_y)\sin(\alpha t),$$
$$A_y(t)=S_y+(A_x(0)-S_x)\sin(\alpha t)+(A_y(0)-S_y)\cos(\alpha t).$$
Alternately, we could look up how to rotate a point around the origin by multiplying by a suitable matrix. In the long run, that is a better way of viewing the matter. The calculation we have done can be viewed as a derivation of that matrix formula.