Show that the points $(9,1), (7,9), (-2,12), (6,10)$ are concyclic. 
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.
 A: The perpendicular bisector of $(9,1)$ and $(7,9)$ is the line $\{(8t+8,2t+5)\mid t\in\Bbb R\}$ and the perpendicular bisector of $(7,9)$ and $(-2,12)$ is the line $\left\{\left(3t+\frac52,9t+\frac{21}2\right)\,\middle|\,t\in\Bbb R\right\}$. They intersect at $C=(0,3)$, which is then the center of the circle passing through $(9,1)$, $(7,9)$ and $(-2,12)$. The distance from $(0,3)$ to $(6,10)$ is $\sqrt{85}$, which the distance from $C$ to the other three points. So, the four points are concyclic. See the picture below.

A:  $\;\;\;\;$ Isosceles trapezoids are cyclic.
A: 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.
A: Here's a completely different approach. We regard your 4 points as complex numbers $$z_1 = 9 + i,\ \  z_2 = 7 + 9i,\ \  z_3 = -2 + 12i, \ \  z_4 = 6 + 10i$$
and now we calculate the cross ratio of these points. It is a basic fact about the cross ratio that the cross ratio is real if and only if the points are either colinear or concyclic. They are clearly not colinear, just by looking at them, so if the cross ratio $$C(z_1, z_2, z_3, z_4) = \frac{(z_3-z_1)(z_4-z_2)}{(z_3 - z_2)(z_4-z_1)}$$
is real, then we're done. Plugging everything in, we have:$$C(z_1, z_2, z_3, z_4) = \frac{((-2+12i)-(9+i))((6+10i)-(7+9i))}{((-2+12i) -(7+9i))((6+10i)-(9+i))}=\frac{(-11+11i)(-1+i)}{(-9+3i)(-3+9i)}$$
which we can simplify down to
$$=\frac{11}{3}\frac{-2i}{-10i}$$ which is evidently real, since the i's cancel.
A: For 4 distinct points $A,B,C,D,$ if  $\,\cos ABC= \frac {(A-B)\cdot (C-B)}{\|A-B\|\cdot \|C-B\|}=-\frac {(A-D)\cdot (C-D)}{\|A-D\|\cdot \|C-D\|}=-\cos ADC$ then the angles $ABC+ADC=\pi,$ which is not possible unless $A,B,C,D$ are concyclic.
