To better understand the physics of a turn, maybe it helps to look at the centripetal force in a different way. This force is actually pulling the aircraft sideways into the new direction of movement. When banking into the turn on an easterly course, the centripetal force will accelerate the aircraft to the north and, towards the end of the turn, will decelerate its easterly speed, such that it will have a northerly speed when the turn is complete. The centripetal force is the means by which ground speed gets converted from a purely easterly speed to a northerly speed.
If you fly with 200 kts in a 50 kts headwind and want to fly a northerly track after finishing the turn, you simply stop turning after having decelerated the aircraft by only 150 kts. You don't fly a quarter circle of 90°, but only one of 75.5°; the angle where the cosine has shrunk to 0.25. After that, your easterly speed component is reduced by 150 kts and your northerly speed component has increased to 193.6 kts. When viewed by another plane flying in the same headwind, you have flown a segment of a circle measuring 75.5°. When seen by an observer on the ground, your ground track is a compressed circle segment.
Starting with 250 kts will make you perform the turn at a higher speed, but with the same technique.
Any mistake until now?
Yes! Your assumption that your ground speed would be 200 kts is wrong. At all times, you will maintain a TAS of 200 kts, and the ground speed after the turn in the first case is only 193.6 kts.
In the second case, you start with a TAS of 250 kts and maintain that over the full turn. Your ground speed after having achieved a northern track is now 242 kts.
The question is how that influence the turning circle?
In which of the two cases the turning circle is flatten/stretched and
in which is circular if any? What is the approximate shape of them?
That depends on the observer. Your ground track is compressed in easterly direction because you started at an easterly speed component of 150 kts but used the aerodynamic forces of 200 kts for turning. You stopped turning after having yawed the airplane by only 75.5°, so you now fly a correction angle of 14.5° to achieve a purely northern track. The observer will see you accelerate from 150 kts ground speed to 193.6 kts ground speed (and from 200 kts to 242 kts in the second case).
The shape of the circle will only look round for an observer flying in the same headwind.
In a tailwind, you would keep turning until you have completed a circle segment of 104.5°, so you can let the centripetal force act longer in easterly direction for a ground speed change of 250 kts in easterly direction. Since it will reduce the aircraft's northerly speed component over the last 14.5°, the centripetal force will slow the northerly speed component to 193.6 kts over the last 14.5° of the turn after it had reached 200 kts after turning 90°. Now the ground track is stretched in easterly direction.
Below I added a graph with three ground tracks and airplane symbols overlaid to show the correction angle at the specific point of the turn. All three cases perform a 90° turn and fly initially parallel to the wind direction. I used 50m/s TAS and ±20m/s wind speed; the axes are in meters. The blue line is the case without wind.
Ground track when turning in wind (own work). A turn from a headwind course takes much less time and azimuth change than one from a tailwind course. Only the ground track without wind is a circle, all other ground tracks are ellipses. In the headwind case the axis parallel to the wind direction is the shorter axis and in the tailwind case it is the longer of the two axes of the ellipse.
If you insist on keeping the ground speed constant, you will need to speed up when turning into the wind rsp. to slow down when turning out of the wind. Your second case with 250 KTAS while heading east and 206 kts when flying a northern track is very unusual. This answer gives you the basic equations. The time to fly a 75.5° turn with 30° bank angle (= 1.15 g) is $$t_{75.5°} = \frac{v\cdot\pi\cdot75.5°}{g\cdot\sqrt{n_z^2-1}\cdot180°}$$
which yields 30.4 s for 250 kts (= 118.6 m/s) and 25.1 s for 206 kts, so you will spend 27.75 s turning when the deceleration is linear. The deceleration along the flight path is 0.455 m/s², so you will need to throttle down during the turn and throttle up when the deceleration phase is over. Good luck keeping the ball centered while turning and changing the p-factor at the same time!
