We have the points $Z=(-1,1)$, $A=(-1,6)$ and $B=(3,4)$.
Let $\delta$ be the rotation with center $Z$ and $\delta (A)=B$.
Let $C$ be the point on the circumcircle of the triangle $ABZ$ such that the segment $\overline{CZ}$ goes through the center of circumcircle.
Let $\gamma$ be a rotation with center $C$ and $\gamma (B)=A$.
Show that $\gamma\circ\delta$ is a point reflection at $A$.
To show that do we have to show that there is exactly one fixed point? Or do we have to determine the matrices of the transformation and consider th determinant of the matrix?
Hint: There is a theorem which says:
If $\mathcal R_M$ and $\mathcal R_N$ are two rotations around $M$ respectively $N$ for $\alpha$ and $\beta$ then their composite $\mathcal R = \mathcal R_N\circ\mathcal R_M$ is rotation around new point $S$ for $\alpha+\beta $ where $S$ is in triangle $MNS$ with $\angle SMN= \alpha/2$ and $\angle MNS = \beta/2$.