Školjka

%V0 A circle $S$ bisects a circle $S'$ if it cuts $S'$ at opposite ends of a diameter. $S_A$, $S_B$,$S_C$ are circles with distinct centers $A, B, C$ (respectively). Show that $A, B, C$ are collinear iff there is no unique circle $S$ which bisects each of $S_A$, $S_B$,$S_C$ . Show that if there is more than one circle $S$ which bisects each of $S_A$, $S_B$,$S_C$ , then all such circles pass through two fixed points. Find these points. Original Statement: A circle $S$ is said to cut a circle $\Sigma$ diametrically if and only if their common chord is a diameter of $\Sigma.$ Let $S_A, S_B, S_C$ be three circles with distinct centres $A,B,C$ respectively. Prove that $A,B,C$ are collinear if and only if there is no unique circle $S$ which cuts each of $S_A, S_B, S_C$ diametrically. Prove further that if there exists more than one circle $S$ which cuts each $S_A, S_B, S_C$ diametrically, then all such circles $S$ pass through two fixed points. Locate these points in relation to the circles $S_A, S_B, S_C.$