The image contains a problem focused on demonstrating that a set of given points are concyclic, meaning they all lie on a common circle.

Here's a breakdown of the approach:

1. General Circle Equation: The equation of a circle is generally given by: $x^2 + y^2 + 2gx + 2fy + c = 0$ Where $g$, $f$, and $c$ are constants to be determined.

2. Given Points: The points provided are $A(1, 2)$, $B(3, -4)$, $C(5, -8)$, and $D(19, 8)$.

3. Substitution into the Circle Equation: To prove the points lie on the same circle, we substitute each point into the circle equation. By substituting the coordinates, a system of linear equations in terms of $g$, $f$, and $c$ is formed.

4. First Substitution (Point $A(1, 2)$): Substitute $(1, 2)$ into the general equation: $1^2 + 2^2 + 2g(1) + 2f(2) + c = 0$ Simplifying this gives: $1 + 4 + 2g + 4f + c = 0$ $2g + 4f + c + 5 = 0 \quad \text{(Equation 1)}$

5. Second Substitution (Point $B(3, -4)$): Substitute $(3, -4)$ into the general equation: $3^2 + (-4)^2 + 2g(3) + 2f(-4) + c = 0$ Simplifying this gives: $9 + 16 + 6g - 8f + c = 0$ $6g - 8f + c + 25 = 0 \quad \text{(Equation 2)}$

6. Third Substitution (Point $C(5, -8)$): Substitute $(5, -8)$ into the general equation: $5^2 + (-8)^2 + 2g(5) + 2f(-8) + c = 0$ Simplifying this gives: $25 + 64 + 10g - 16f + c = 0$ $10g - 16f + c + 89 = 0 \quad \text{(Equation 3)}$

Next Steps:

You would typically solve this system of linear equations (Equations 1, 2, and 3) for $g$, $f$, and $c$. After solving these, if the constants satisfy the equation when the fourth point $D(19, 8)$ is substituted, all points are concyclic.

Would you like help solving the system of equations?