In the Klein model, one may see that this is also a circle. Consider a line segment with one point on the center of the disk. One side of the triangle goes through the center. Then orthogonal lines to a line through the center are also orthogonal in the hyperbolic metric, e.g. since they are preserved by reflection. So one sees that a circle is traced out which goes through the origin. If you'd rather center the curve at the origin, then it will be an ellipse, since hyperbolic isometries of the Klein model are projective transformations.
To convert to the Poincare model, take a hemisphere sitting over the disk, and project vertically. The projection of the circle is given by the intersection of a cylinder over the circle with the upper hemisphere. This upper hemisphere is conformally equivalent to the Poincare model, e.g. by inversion through a sphere centered at the south pole of the lower hemisphere. I haven't computed the curve this traces out though.
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