An ellipse of major axis 20√3 and minor axis 20 slides along the coordinate axes and always remain confined in the 1st quadrant

An ellipse of major axis $20√3$ and minor axis $20$ slides along the coordinate axes and always remain confined in the 1st quadrant. The locus of the center of ellipse therefore describes an arc of circle. The length of this arc is $\dots$

Attempt:- Let the foci of the required ellipse be $(X1,Y1)$ & $(X2,Y2)$. Now since it slides along the coordinate axes, the axes are tangents to the required ellipse.

Product of perpendiculars from foci to any tangent is $b²$.

Relevant equations:- center is at {$h=(X1+X2)/2$, $k=(Y1+Y2)/2$},

Distance between foci is $\sqrt{(X1-X2)^2 + (Y1-Y2)^2}=4a^2e^2$

$X1X2=Y1Y2=b^2$

Hence locus of the center is of the ellipse is $X^2 + Y^2 = 400$. But how to find the length of the arc?

The nontrivial thing here is the fact that the arc in question is indeed circular (I didn't know this). Believing that we can take a look at the following figure where the ellipse stays fixed and the angle formed by the positive $x$- and $y$-axes rotates. As the semiaxes of the ellipse are $10\sqrt{3}$ and $10$ the radius of the arc in question is $20$. The black angle in the figure is $30^\circ$, and during the rotation it increases to $60^\circ$. It follows that our arc has length $20\cdot{\pi\over6}={10\pi\over3}$. This arc is covered four times during a full $2\pi$-turn of the ellipse vs. the angle.
When the ellipse is perfectly vertical, its axes will be aligned with the co-ordinate axes and the line joining its centre and the origin of co-ordinates will make an angle of $60^\circ$ with the X-axis. This is because the tangent of the angle will be $\tan\theta=\dfrac{10\sqrt 3}{10}=\sqrt 3$.
Similarly, when the ellipse becomes perfectly horizontal the angle will be $30^\circ ,\implies$that the centre travels only $30^\circ$ on the circle you obtained as the locus. The arc length can now be easily obtained.