we have to show that four points (-6, 0), (-2, 2) , (-2,-8) and (1 ,1) are concyclic.

SOLUTION : four points (-6, 0), (-2, 2) , (-2,-8) and (1 ,1) to be concyclic only if they should lie on a circle.

let a circle of EQUATION is ...

x² + y² + 2gx + 2fy + C = 0

if (-6,0) lies on the circle it should satisfy the equation.

(-6)² + 0² + 2g(-6) + 2f(0) + c = 0

⇒36 - 12g + c = 0 .........(1)

similarly, (-2, 2) lies on the circle,

so, (-2)² + 2² + 2g(-2) + 2f(2) + c = 0

8 - 4g + 4f + c = 0 ........(2)

and (-2,-8) lies on the circle it should satisfy the equation.

(-2)² + (-8)² + 2g(-2) + 2f(-8) + c = 0

⇒68 - 4g - 16g + c = 0 ........(3)

after solving EQUATIONS (1), (2) and (3) we get

c = -12, g = 2 and f = 3

so the equation is ...

x² + y² + 4x + 6y - 12 = 0..........(4)

here the fourth point should automatically satisfy this equation for the points to be concyclic.

now putting (1,1) in equation (4),

(1)² + (1)² + 4(1) + 6(1) - 12 = 12 - 12 = 0

Therefore four points (-6, 0), (-2, 2) , (-2,-8) and (1 ,1) are concyclic.