Given a triangle of vertices $A1(4,0), A2(-4c, 0)$ and $A3(4,-c)$, where $c > 0$, write the equation of its nine-point circle and compute the nine points.
Solution Let $L, M$ and $N$ be the midpoints of $A1A2$. $A1A3$ and $A2A3$: these are three of the nine points of the nine point circle and completely determine it. A computation of the midpoint shows that $L = (2 - 2c, 0), \, M = (4, - c / 2)$ and $N = (2 - 2c, - c / 2)$
The center of the nine point circle is the circumcenter of the triangle ALMN: since the sides $LV$ and $MN$ are respectively vertical and horizontal, their perpendicular bisectors are horizontal and vertical, and so easily determined by the passage through their midpoints, namely, $y = -C/4$ and $x = 3 - c$
Therefore, the center $(x-(3-c))^2$ + $(y+c/4)^2$ of the nine point circle. The radius $\sqrt{(1+c)^ 2 + c^ 2} = 16$ .
How do we compute the nine points?