Knowing the equation of the tangent to a rotated curve, how to find the point of tangency.?

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Note : in the case of non rotated curves, the coordinates of the point of tangency can be " read off" on the equation of the tangent ( point-slope formula ... ) , but it's not the case here; hence my question

Let $C$ be the curve defined by : $f(x)= 3 \cos(x/3)$.

Let $(a, f(a))$ be a moving point on $C$.

Let $C'$ be the image of $C$ under a counterclockwise rotation of $R$ radians; namely, $C'$ is defined by :

$ Y(x,y) - f( X(x,y)) = 0$ ( or, equivalently : $ Y(x,y) = f( X(x,y)))$,

with ( by the rotation of axis formula)

$X(x,y) = x\cos(R) + y \ sin(R)$

and

$Y(x,y)= y \cos(R) - x \sin(R) $.

Let
$$X(x,y) - f(a) = f'(a) ( X(x,y) -a )$$

be a moving straight line constantly tangent to $C'$.

My question is how to define the ( moving) point of tangency ( or the moving point of contact between $C'$ and its moving tangent) , in terms of $a$.

Note : The attenpt at defining the point of tangency as $( X(a, f(a)), Y(a, f(a)) $ does not work, because this last point follows the image of the curve rotated in the wrong sense.