The question is asking to show that the points $(1, 2), (3, -4), (5, -6), (19, 8)$ are concyclic. This means that these points lie on the same circle.

Steps to prove that four points are concyclic:

1. Condition for Concyclic Points: Four points $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3), D(x_4, y_4)$ are concyclic if the determinant of the following matrix is zero: x_1 & y_1 & x_1^2 + y_1^2 & 1 \\ x_2 & y_2 & x_2^2 + y_2^2 & 1 \\ x_3 & y_3 & x_3^2 + y_3^2 & 1 \\ x_4 & y_4 & x_4^2 + y_4^2 & 1 \end{vmatrix} = 0$$ This determinant is known as the **cyclic determinant**.
2. Substitute the Coordinates: The given points are $A(1, 2), B(3, -4), C(5, -6), D(19, 8)$. We substitute the coordinates into the matrix: 1 & 2 & 1^2 + 2^2 & 1 \\ 3 & -4 & 3^2 + (-4)^2 & 1 \\ 5 & -6 & 5^2 + (-6)^2 & 1 \\ 19 & 8 & 19^2 + 8^2 & 1 \end{vmatrix} = 0$$ Simplify the squares: $$\begin{vmatrix} 1 & 2 & 5 & 1 \\ 3 & -4 & 25 & 1 \\ 5 & -6 & 61 & 1 \\ 19 & 8 & 425 & 1 \end{vmatrix}$$
3. Evaluate the Determinant: Now, we evaluate the determinant of this 4x4 matrix to check if it is zero. If the determinant is zero, the points are concyclic.

I'll compute the determinant for you next.The determinant of the matrix is approximately $9.33 \times 10^{-13}$, which is extremely close to zero. This indicates that the points $(1, 2), (3, -4), (5, -6), (19, 8)$ are indeed concyclic, as the determinant is essentially zero.

Would you like more details on any step? Here are five related questions to consider:

1. What is the geometric significance of concyclic points?
2. Can three points always be considered concyclic? Why or why not?
3. How is the equation of a circle determined from three points?
4. How does this method change if we are testing for more than four points?
5. Can concyclic points also form a regular polygon? Under what conditions?

Tip: Concyclic points always share a common circumcircle. If you're ever in doubt, calculating the cyclic determinant as we've done is a reliable method to check!