Pedal circle



The pedal circle of the a triangle $$ABC$$ and a point $$P$$ in the plane is a special circle determined by those two entities. More specifically for the three perpendiculars through the point $$ P$$ onto the three (extended) triangle sides $$ a,b,c$$ you get three points of intersection $$P_a, P_b, P_c$$ and the circle defined by those three points is the pedal circle. By definition the pedal circle is the circumcircle of the pedal triangle.

For radius $$r$$ of the pedal circle the following formula holds with $$R$$ being the radius and $$O$$ being the center of the circumcircle:
 * $$r=\frac{|PA| \cdot |PB| \cdot |PC|}{2\cdot (R^2-|PO|^2)} $$

Note that the denominator in the formula turns 0 if the point $$P$$ lies on the circumcircle. In this case the three points $$P_a, P_b, P_c $$ determine a degenerated circle with an infinite radius, that is a line. This is the Simson line. If $$P$$ is the incenter of the triangle then the pedal circle is the incircle of the triangle and if $$P$$ is the orthocenter of the triangle the pedal circle is the nine-point circle.

If $$P$$ does not lie on the circumcircle then its isogonal conjugate $$Q$$ yields the same pedal circle, that is the six points $$P_a, P_b, P_c $$ and  $$Q_a, Q_b, Q_c $$ lie on the same circle. Moreover, the midpoint of the line segment $$PQ$$ is the center of that pedal circle.

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.

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 $$ABC$$ and the fourth point as the point $$P$$, then they define a pedal circle. The four pedal circles you get this way intersect in a common point.