Three circles are given $k_1$,$k_2$,$k_3$ that have two common points A and B. Prove that every circle $k$ that is orthogonal to circles $k_1$,$k_2$, is also orthogonal to $k_3$.
Here is my proof although i don;t know if it is correct. It was too easy so i'm skeptical.
Lets pic $n(A,r)$ be arbitrary and let $\Psi_n$ be inversion with respect to circle $n$. Let $\Psi_n$:$k_1, k_2,k_3,k \rightarrow c_1, c_2,c_3,c $ Since $A \in k_1, k_2,k_3$, $\Psi_n$ maps them into lines. And since $B \in k_1, k_2,k_3$ it follows that $c_1, c_2,c_3$ intersect in $B'$(where $B'$ is map of $B$). Since $k$ is orthogonal to $k_1$ and $k_2$ and since inversion preserves angles it follows that $c_1$, $c_2$ are orthogonal to $c$. Hence it follows that both $c_1$ and $c_2$ pass through center of circle $c$ and lets call that center of circle $C$.Since we cont that $c_1$ and $c_2$ intersect at $B'$ and at the same time intersect at $C$ it follows that $C=B'$. We know that $c_3$ goes through $B'$ i.e. it goes through $C$ the center of circle c. So we have that $c_3$ is orthogonal to circle c. Since inversion preserves angles it follows that $k_3$ and $k$ are orthogonal.
Yes, you consider a transformation taking circles to circles ( or lines) and taking one of the points to infinity. The three circles become three lines through the image of the other point. Note that your transformation preserves the angles. Now it's easy. Exactly what you did.