in the plane is a special circle determined by those two entities.
More specifically for the three perpendiculars through the point
onto the three (extended) triangle sides
of the pedal circle the following formula holds with
being the center of the circumcircle:[2] Note that the denominator in the formula turns 0 if the point
determine a degenerated circle with an infinite radius, that is a line.
does not lie on the circumcircle then its isogonal conjugate
yields the same pedal circle, that is the six points
Moreover, the midpoint of the line segment
is the center of that pedal circle.
[1] Griffiths' theorem states that all the pedal circles for a points located on a line through the center of the triangle's circumcircle share a common (fixed) point.
[4] Consider four points with no three of them being on a common line.
Then you can build four different subsets of three points.
Take the points of such a subset as the vertices of a triangle
The four pedal circles you get this way intersect in a common point.