To prove that the points $(1, 1)$, $(-6, 0)$, $(-2, 2)$, and $(-2, -8)$ are concyclic, we need to show that they lie on the same circle. This requires confirming that the perpendicular bisectors of the chords formed by any two pairs of points meet at a single point, which will be the center of the circle. We can then find the equation of the circle.

Step 1: The equation of a circle

The general equation of a circle in 2D is: $(x - h)^2 + (y - k)^2 = r^2,$ where $(h, k)$ is the center of the circle, and $r$ is its radius.

Step 2: Perpendicular bisectors of two chords

We select two chords:

1. Between $A(1, 1)$ and $B(-6, 0)$.
2. Between $A(1, 1)$ and $C(-2, 2)$.

Finding the perpendicular bisector of $AB$:

The midpoint of $AB$ is: $M_{AB} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{1 - 6}{2}, \frac{1 + 0}{2}\right) = (-2.5, 0.5).$ The slope of $AB$ is: $\text{slope of } AB = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 1}{-6 - 1} = \frac{-1}{-7} = \frac{1}{7}.$ The slope of the perpendicular bisector is the negative reciprocal: $\text{slope of perpendicular bisector} = -7.$ The equation of the perpendicular bisector passing through $(-2.5, 0.5)$ is: $y - 0.5 = -7(x + 2.5).$ Simplify: $y = -7x - 17.$

Finding the perpendicular bisector of $AC$:

The midpoint of $AC$ is: $M_{AC} = \left(\frac{1 - 2}{2}, \frac{1 + 2}{2}\right) = (-0.5, 1.5).$ The slope of $AC$ is: $\text{slope of } AC = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 1}{-2 - 1} = \frac{1}{-3} = -\frac{1}{3}.$ The slope of the perpendicular bisector is the negative reciprocal: $\text{slope of perpendicular bisector} = 3.$ The equation of the perpendicular bisector passing through $(-0.5, 1.5)$ is: $y - 1.5 = 3(x + 0.5).$ Simplify: $y = 3x + 3.$

Step 3: Finding the intersection of the perpendicular bisectors

Solve the equations of the two perpendicular bisectors:

1. $y = -7x - 17$,
2. $y = 3x + 3$.

Equating $-7x - 17 = 3x + 3$: $-7x - 3x = 3 + 17,$ $-10x = 20 \implies x = -2.$ Substitute $x = -2$ into $y = 3x + 3$: $y = 3(-2) + 3 = -6 + 3 = -3.$ The center of the circle is $(-2, -3)$.

Step 4: Radius of the circle

The radius $r$ is the distance from the center $(-2, -3)$ to any of the points, say $A(1, 1)$: $r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(1 - (-2))^2 + (1 - (-3))^2} = \sqrt{(1 + 2)^2 + (1 + 3)^2}.$ $r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$

Step 5: Equation of the circle

The equation of the circle is: $(x + 2)^2 + (y + 3)^2 = 25.$

Final Answer:

The points are concyclic, and the equation of the circle is: $(x + 2)^2 + (y + 3)^2 = 25.$

Would you like further clarification or derivations?

---

1. How do we determine if points are concyclic geometrically?
2. Can the method of circumcenter finding be applied in 3D space?
3. What are the other ways to verify that points are concyclic?
4. How can a determinant method be used to prove concyclicity?
5. How would you use computational tools to solve this?

Tip: The determinant method involves verifying if the determinant of a special 4x4 matrix (constructed with the points' coordinates) is zero for concyclic points.