Drag the sliders to explore triangles with vertices on the unit circle and their nine-point circles. Use the bookmark for right triangles, moving the slider.

Three points on the unit circle can be specified by their polar angles:

.

The vertices of triangle , then, are at the complex points

.

The midpoints (red) of the sides , , are simply

.

These three points define the nine-point circle, which has its center at

and radius . The formula for the nine-point circle is

.

The points (blue) for the intersections of the altitudes on the extended baselines {AB, AC, BC} are

.

The orthocenter (point O), which is the intersection of the extended altitudes of the triangle, is

.

The final three points (green) on the nine-point circle are the midpoints of the segments from the vertices to the orthocenter:

.

Thus we have a complete solution for the nine-point circle. From this discussion, it is apparent that there exist solutions for degenerate triangles where two or all of the points are the same, yielding a line segment or a point. To show the degenerate solutions, set two or all three of the sliders to the same value.