Circle inversion of a circle Given is a circle K with radius r and centre M1. K' is a second circle with radius r' and centre M2 that cuts K in two points A and B so that $[M1A]$ is orthogonal to $[M2A]$ and also $[M1B]$ is orthogonal to $[M2B]$. 
We noted:


*

*Now through inversion on K, every circle K' is being reflected onto itself. 


Why is this so?
We also noted that a circle conversion of a line gives a line again. Again: why a line and not a circle? Trying to visualize it I get an arc..

 A: I think this is the situation:


As you can see the inversion of any point C on K' is a point F which is also on K'.
A: Regarding a proof of the proposition, consider the following diagram:

So we are given:


*

*The triangle formed by the two centres and $A$ is right. Algebraically $d^2=r^2+r'^2$.

*$C$ is a random point on the second circle with inversion $C'$. $D$ is their midpoint. So $C,C'$ lie at distances $q+t,q-t$ from $M_1$ respectively.

*$C,C'$ being inversions means that $(q+t)(q-t)=r^2$.


We are two show that:


*

*$C'$ lies on the blue circle iff $C$ does.

*This is equivalent to saying that the midpoint $D$ is placed in a way so that the dotted line $s$ meets the segment $C' C$ at a right angle.

*In other words, $s$ must divide the triangle formed by the two centres and the point $C$ into two right triangles.

*So if we can show that $d^2-q^2=s^2=r'^2-t^2$, we are done.


The steps to show this are:


*

*From earlier on we have: $r^2=(q+t)(q-t)=q^2-t^2$.

*Substituting this expression for $r^2$ into the equation $d^2=r^2+r'^2$ and subtracting $q^2$ from both sides leads to the desired goal.

