Show that, given four coplanar points, we can always draw two intersecting circles coplanar with the points, such that two of the given points are diameter endpoints of one circle, and the other two given points are diameter endpoints of the other circle.
In this question, "intersecting" means that the circles share at least one common point.
Example:
I will post my answer. Alternative solutions are welcome.
This question and answer serve to provide ideas that might help answer a harder question about five points and two non-intersecting circles.
Best Answer
One can also do it with vector algebra. Note that the circles with diameters $A,B$ and $C,D$ do not intersect if and only if $$ |(A-C)\cdot(B-D)+(A-D)\cdot(B-C)|>|A-B||C-D|\,. $$ Now out of 3 possible sums of scalar products of this type inside the absolute value, two have the same sign. Without loss of generality those are the one written above and $$ (A-B)\cdot(C-D)+(A-D)\cdot(C-B) $$ Then the sum of their absolute values is the absolute value of the sum, which should be larger than the sum of the corresponding right hand sides if there are no intersections. But adding honestly, we get $$ |(A-C)\cdot(B-D)+(A-B)\cdot(C-D)|>|A-B||C-D|+|A-C||B-D|\, $$ which is absurd.
I tried to apply this technique to the 5-point question but without success so far. The difficulty is that now the scalar products have to be estimated by absolute values from below, which makes Cauchy-Schwarz hard to use in a meaningful way.