How is the determinant related to the unit circle? Here is a STEP 3 question from 2018, of which the Latex code can be downloaded here. I know perfectly how to answer this question, so please do not answer it.

The  distinct points $A$, $Q$ and $C$ lie on a straight line in the
Argand diagram, and  represent the distinct complex numbers $a$, $q$
and $c$, respectively.  Show that $\dfrac {q-a}{c-a}$  is real and
hence that  $(c-a)(q^*-a^*) = (c^*-a^*)(q-a)\,$.
Given that $aa^* = cc^* = 1$, show further that  $$ q+ ac q^* = a+c
 $$
The distinct points $A$, $B$, $C$ and $D$ lie, in anticlockwise order,
on the circle of unit radius  with centre at the origin   (so that,
for example, $aa^* =1$). The lines $AC$ and $BD$ meet at $Q$. Show
that  $$ (ac-bd)q^* = (a+c)-(b+d) \,, $$ where   $b$ and $d$  are
complex numbers represented by the  points $B$ and $D$   respectively,
and show further that $$ (ac-bd)  (q+q^*) =  (a-b)(1+cd) +(c-d)(1+ab)
 \,. $$
The lines $AB$ and $CD$ meet at $P$, which  represents the complex
number $p$. Given that~$p$ is real, show that $p(1+ab)=a+b\,$. Given
further that $ac-bd \ne 0\,$, show that  $$ p(q+q^*) =  2  \,. $$

This question involves a large amount of algebra, and one can easily get lost in it. However, there are some interesting patterns emerging in the problem:

*

*Why $ac-bd$ appears? This is the $2\times 2$ determinant! What's really going on here? In complex analysis/methods books, I have not seen any determinants related to circles or straight lines. (For circles, we have cross ratios, which is something completely different.)

*Are there any generalizations to $3\times 3$ or higher determinants?

*The last result, which states that $p\Re (q)=1$, is also quite interesting. How to interpret this result?

To summarize, this question seems to be tackling a very deep theory in geometry in an elementary way. I try to find more about this in complex methods textbooks, but I have not found any.
Can anyone explain intuitively what's going on about the determinants, or tell me where to read more about the geometry of the unit circle?
Just tell me where this question leads us to if we go any further.
 A: A  very partial answer: To me a determinant is related to some linear structure, and in the present situation I do not see any underlying linear structure giving rise to $ac-bd$ in a natural way.
In view of the lack of replies I imagine that I am not the only one having this problem.
The expression is intimately related to the fact that we look at points on the unit circle. For such points it is, however, natural that we obtain an expression for $q^*$ which becomes singular when $ac=bd$:
Consider the collection ${\cal D}$ of straight lines intersecting the unit circle $S^1$.
Say that lines in ${\cal D}$ are equivalent if parallel. Let $\ell$ intersect $S^1$
in the points $a$ and $c$ (which are the same when $\ell$ is tangent to $S^1$). Then the mapping
$$ \ell \in {\cal D} \mapsto ac \in S^1$$
induces a bijection between equivalence classes of parallel lines and $S^1$. For example, a translation of $\ell$ corresponds to a shift $a\mapsto a e^{i \phi}$ and $c\mapsto c e^{-i\phi}$, leaving the product invariant. In other words if $\ell'$ intersects $S^1$ in $b$ and $d$ then $\ell\sim \ell'$ iff $ac-bd=0$. When lines $\ell,\ell'\in {\cal D}$ are not identical  then they have a unique intersection point on the Riemann sphere $\widehat{\Bbb C}$, with $\infty$ being reached when the lines are parallel. The formula $q^*=\frac{a+c-b-d}{ac-bd}$ only fails  when the two lines are identical and we have $\frac00$ on the RHS.
There are other nice properties, like e.g. the involution given by the Möbius transformation $M_q(z)= \frac{q-z}{1-q^* z}$ which interchanges $a\leftrightarrow c$, $b\leftrightarrow d$ and the fact that $p(q+I(q))=2$ holds  in general with $I$ being the involution in the line through the origin and $p$ (for general $p$). But this is somewhat off the track as I don't have geometric explanation for this.
