Circle inversion of a circle

Question

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..

Answer 1

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.

Answer 2

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'.