The conjecture is true, and can be recovered from the standard
Sylverster-Gallai theorem in the plane.
We first prove:
Lemma 1 Suppose $C$ is a finite configuration of points
in real 3-space ${\bf R}^3$ such that for every $Q_1,Q_2,Q_3 \in C$
there is a plane meeting $C$ in $Q_1,Q_2,Q_3$, and at least one more
point of $C$. Then $C$ is contained in a plane.
(Of course the plane is unique unless $Q_1,Q_2,Q_3$ are collinear.)
Proof of Lemma 1: fix $Q_1 \in C$ and let $\Pi_1$ be a plane
containing $Q_1$ but no other point of $C$. Let $\Pi \neq \Pi_1$
be any plane parallel to $\Pi$, and let $C'$ be the set consisting
of the projections from $Q_1$ to $\Pi$ of all $Q \in C$ other than $Q_1$
(that is, the intersections with $\Pi$ of the lines $\overline{Q_1 Q}$).
Applying the hypothesis only to triples that contain $Q_1$
shows that $C'$ satisfies the hypothesis of the Sylvester-Gallai theorem.
Hence $C'$ is contained in a line, whence $C$ is contained in the plane
spanned by this line and $Q_1$. $\ \diamondsuit$
We connect this with the problem at hand using the following observation:
Lemma 2 Points $(x_i,y_i)$ in the plane are concyclic or collinear
iff the corresponding points $(x_i, y_i, x_i^2 + y_i^2)$ on the
round paraboloid $z = x^2 + y^2$ are coplanar.
Proof of Lemma 2: the intersection of $z = x^2 + y^2$ with any plane
$A_0 + A_1 x + A_2 y + A_3 z = 0$ projects to the locus of
$A_0 + A_1 x + A_2 y + A_3 (x^2+y^2) = 0$, which is line if $A_3 = 0$
and a circle otherwise. $\ \diamondsuit$
Now assume that $S \subset {\bf R}^2$ is a finite set of points
such that for every $P_1,P_2,P_3 \in S$ there is a circle meeting $S$ in
$P_1,P_2,P_3$, and at least one more point of $S$. Then by Lemma 2
the associated configuration
$$
C_S := \{ (x,y,x^2+y^2) \in {\bf R}^3 \mid (x,y) \in S \}
$$
satisfies the hypotheses of Lemma 1. Hence $C_S$ is contained in a plane.
Applying Lemma 2 in reverse, we conclude that $S$ is contained in a circle
or line. The line does not satisfy the hypothesis, so $S$ is concyclic,
as desired. QED
The same linearization trick applies to other such problems.
Lemma 1 generalizes to ${\bf R}^d$ for any $d>2$: a finite configuration $C$
such that every $d$ points are on a hyperplane through a $(d+1)$st point
must be contained in a hyperplane; this is proved by projection to a plane
from any $d-2$ points of $C$ in general linear position. So, for instance,
if $S \subset {\bf R}^2$ has the property that every five points are
on a conic that contains a sixth point of $S$ then the associated
configuration
$$
\{ (x,y,x^2,xy,y^2) \in {\bf R}^5 \mid (x,y) \in S \}
$$
(which lies on the affine version of the
Veronese surface)
is contained in a hyperplane, whence $S$ is contained in a conic.
$AHC$ is a right triangle and the median relative to hypotenuse is half the hypotenuse
$AH=\frac12 FD\to AH=AD$
$ADCB$ is a parallelogram so AD=BC$
therefore $AH=BC$
Furthermore $AB\parallel DE$ and this proves that $ABCH$ is an isosceles trapezoid which is always possible to inscribe in a circle.
Hope this helps
EDIT
The median $AH$ is half $DF$ because you can build a rectangle as I did in the second picture, diagonals are equal and bisect each other.
$$...$$
Best Answer
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.