In a solution to the problem, IMO 2015 #3 Evan Chen and Telv Cohl used the notation of negative inversion which looks like Homothety:
Here , $\angle$$AML$=$\angle ABL=\angle AKL$.
Also,$H, J$ are the midpoints of $BC, AC$.$D, E, F$ are the feet of the perpendiculars drawn from$C, A, B$ to their opposite sides.$I$ is the orthocenter of this triangle.
Now, if I perform a negative inversion , then what will happen to the circles passing through the center? also, what will happen to the lines not passing through the center?
Can this inversion take the $\odot(ABC)$ to its nine-point circle? As the plain inversion around the circle with center$O$ sends point $P$ to $P'$ so that $OP$$\times$$OP'=r^2$ where r is the radius of the circle we are inverting around.
How do actually negative inversion map the points?
Negative inversion is inversion in a circle with an imaginary radius. For example, the circle $x^2+y^2=-r^2$ has radius $ir$, where $i=\sqrt{-1}$.
Such "imaginary" circles exist, in the same sense that the "imaginary" number $i$ exists, and you can work with them using the same algebra as for real circles.
Whereas inversion in a real circle with radius $r$ takes $P\to P'$ so that $OP\times OP'=r^2$, inversion in an imaginary circle satisfies $OP\times OP'=-r^2$. i.e. $P$ and $P'$ are in opposite directions from center $O$.
It follows that negative inversion is "regular" inversion followed by a reflection in $O$.
See also cut-the-knot