214. Sliding Ellipse. Let two mutually perpendicular lines be given, and let some ellipse move in the plane, constantly touching both lines. Determine what line the center of the ellipse will describe during this motion.
214. Consider the canonically positioned ellipse $b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}$. The geometric locus of points of intersection of two mutually 

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perpendicular tangents to this ellipse is the circle $x^{2} + y^{2} = a^{2} + b^{2}$. From this, it follows that if we fix two mutually perpendicular lines and move the ellipse along the plane so that it touches these lines, the center of the ellipse will always be at a distance of $\left(a^{2} + b^{2}\right)^{\frac{1}{2}}$ from the point of intersection of these lines. If we take these two lines as the coordinate axes, then the center of our ellipse will be closest to these axes at the points $(a, b)$ and $(b, a)$. Therefore, as the ellipse moves, its center will trace the smaller arc of the circle $x^{2} + y^{2} = a^{2} + b^{2}$, enclosed between these points. The length of such an arc is

$$
\left(a^{2} + b^{2}\right)^{\frac{1}{2}} \operatorname{arctg} \frac{a^{2} - b^{2}}{2 a b}
$$

[A. Struik, M. M., 24, 231 (March 1951).]