Pedal circle

${\displaystyle \triangle ABC}$ with sides ${\displaystyle a,b,c}$ and point ${\displaystyle P}$
feet of the perpendicular: ${\displaystyle P_{a},P_{b},P_{c}}$
center of the circumcircle: ${\displaystyle O}$
the green segments are used in the formula for radius

${\displaystyle \triangle ABC}$ with isogonal conjugates ${\displaystyle P}$ and ${\displaystyle Q}$
6 feet on the pedal circle: ${\displaystyle P_{a},P_{b},P_{c},Q_{a},Q_{b},Q_{c}}$
center of the pedal circle and midpoint of ${\displaystyle PQ}$ : ${\displaystyle M}$
angle bisectors: ${\displaystyle w_{a},w_{b},w_{c}}$

4 points ${\displaystyle A,B,C,D}$ and 4 pedal circles intersecting in ${\displaystyle S}$

The pedal circle of the a triangle ${\displaystyle ABC}$ and a point ${\displaystyle P}$ in the plane is a special circle determined by those two entities. More specifically for the three perpendiculars through the point ${\displaystyle P}$ onto the three (extended) triangle sides ${\displaystyle a,b,c}$ you get three points of intersection ${\displaystyle P_{a},P_{b},P_{c}}$ and the circle defined by those three points is the pedal circle. By definition the pedal circle is the circumcircle of the pedal triangle.^[1][2]

For radius ${\displaystyle r}$ of the pedal circle the following formula holds with ${\displaystyle R}$ being the radius and ${\displaystyle O}$ being the center of the circumcircle:^[2]

${\displaystyle r={\frac {|PA|\cdot |PB|\cdot |PC|}{2\cdot (R^{2}-|PO|^{2})}}}$

Note that the denominator in the formula turns 0 if the point ${\displaystyle P}$ lies on the circumcircle. In this case the three points ${\displaystyle P_{a},P_{b},P_{c}}$ determine a degenerated circle with an infinite radius, that is a line. This is the Simson line. If ${\displaystyle P}$ is the incenter of the triangle then the pedal circle is the incircle of the triangle and if ${\displaystyle P}$ is the orthocenter of the triangle the pedal circle is the nine-point circle.^[3]

If ${\displaystyle P}$ does not lie on the circumcircle then its isogonal conjugate ${\displaystyle Q}$ yields the same pedal circle, that is the six points ${\displaystyle P_{a},P_{b},P_{c}}$ and ${\displaystyle Q_{a},Q_{b},Q_{c}}$ lie on the same circle. Moreover, the midpoint of the line segment ${\displaystyle PQ}$ 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 ${\displaystyle ABC}$ and the fourth point as the point ${\displaystyle P}$ , then they define a pedal circle. The four pedal circles you get this way intersect in a common point.^[3]

References

1. Ross Honsberger: Episodes in Nineteenth and Twentieth Century Euclidean Geometry. MAA, 1995, pp. 67–75
2. Roger A. Johnson: Advanced Euclidean Geometry. Dover 2007 (reprint), ISBN 978-0-486-46237-0, pp. 135–144, 155, 240
3. Weisstein, Eric W. "Pedal Circle". MathWorld.
4. Weisstein, Eric W. "Griffiths' Theorem". MathWorld.

External links

• Media related to Pedal circle at Wikimedia Commons
• Pedal Circle of Isogonal Conjugates - interactive illustration in GeoGebra
• pedal triangle and pedal circle - interactive illustration