show that the Four Points (- 6, 0 ) (- 2, 2) (- 2, - 8) and (1,1)are concyclic
Answers
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.
Answer:
show that the four points (1,1),(-6,0),(-2,2),(-2,-8) are concyclic