TITLE: Show that the points $(9,1), (7,9), (-2,12), (6,10)$ are concyclic.
QUESTION [6 upvotes]: Show that the points $(9,1), (7,9), (-2,12), (6,10)$ are concyclic.

How can we prove that the given points are con-cyclic?
I know the fact the points are said to be concyclic if they lie on the same circle. I substituted the coordinates in the equation of circle and got 4 equations:-

$(9-h)^2 + (1-k)^2 = r^2$
$ (7-h)^2 + (9-k)^2 = r^2$
$(-2-h)^2 + (-12-k)^2 = r^2$
$(6-h)^2 + (10-k)^2 = r^2$

Now, here picking the first three equations, I got the centre of circle as $(-8,1)$ and radius = $17$units.
I'm getting no idea what to do further. Is there any short method to solve the question? Please help me here.

REPLY [3 votes]: You can use Ptolemy's inequality to check if the points are concyclic.
Define $A=(9,1),B=(7,9),C=(-2,12),$ and $D=(6,10)$. The inequality states that the points $A$, $B$, $C$, and $D$ are concyclic if $AB \cdot CD + AD 
\cdot BC = AC + BD$.
By solving for the distances, we have $AB=2\sqrt{17},CD=2\sqrt{17}, AC=11\sqrt{2}, BD=\sqrt{2}, AD=3\sqrt{10}$, and $BC=3\sqrt{10}$. You can check that the inequality becomes an equality, which means they are concyclic.