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.. enter image description here

  • $\begingroup$ This is completely wrong. $\endgroup$ – darij grinberg Apr 8 '16 at 13:16
  • $\begingroup$ Inversion with which centre? which radius? $\endgroup$ – Bernard Apr 8 '16 at 13:19
  • $\begingroup$ Inversion with centre of K. More it does not say. $\endgroup$ – Vazrael Apr 8 '16 at 13:41
  • $\begingroup$ By middlepoints, you mean centers of the circles? $\endgroup$ – Jean Marie Apr 8 '16 at 13:41
  • $\begingroup$ @JeanMarie Yes, exactly. I should correct this above! $\endgroup$ – Vazrael Apr 8 '16 at 13:52

Regarding a proof of the proposition, consider the following diagram:

enter image description here

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.
  • $\begingroup$ Awesome! Thank you so much :) $\endgroup$ – Vazrael Apr 9 '16 at 8:52

I think this is the situation:

pic1 pic2

As you can see the inversion of any point C on K' is a point F which is also on K'.

  • $\begingroup$ Thank you! How did you do it? Would you mind sending me the ggb-data? :) $\endgroup$ – Vazrael Apr 8 '16 at 16:14
  • 1
    $\begingroup$ Sorry, I did not save it. I created the red circle, then added a point outside of it for $M_2$. I took the tangent lines from that point and marked the tangent points $A$ and $B$. The blue circle goes through those points. Tangent lines are orthogonal to radial lines. To get the green points, I picked an arbitrary point on the blue circle and drew tangent lines to the red circle. Finally the inversion is on the midpoint between $D$ and $E$. $\endgroup$ – ja72 Apr 8 '16 at 17:03

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