Question

show that the points (1,1)(-6,0),(-2,2) and (-2,-8) are concyclic and find the equation of the circle​

Answer 1

Concept:

If a set of points is on a common circle, they are said to be concyclic (or concyclic) in geometry. The distance between all concyclic points and the circle's center is the same.

Given:

Four points are given to us (1,1)(-6,0),(-2,2) and (-2,-8)

To find:

show that the points (1,1), (-6,0), (-2,2,), and (-2,-8) are concyclic and find the equation of the circle​.

Solution:

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

Let the equation of circle be

$x^{2} + y^{2} + 2gx + 2fy + c = 0$

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

$(-6)^{2} + 0 + 2g(-6) + 2f(0) + c = 036 - 12g + c = 0$ .........(1)

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

So,

$(-2)^{2} + 2^{2} + 2g(-2) + 2f(2) + c = 08 - 4g + 4f + c = 0$........(2)

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

$(-2)^{2} + (-8)^{2} + 2g(-2) + 2f(-8) + c = 068 - 4g - 16g + c = 0$........(3)

After solving equations (1), (2), and (3) we get

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

So the equation is ...

$x^{2} + y^{2} + 4x + 6y - 12 = 0........$..(4)

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

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

$(1)^{2} + (1)^{2} + 4(1) + 6(1) - 12 = 12 - 12\\=0$

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

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Answer 2

Answer:

The given four points are con-cylic.

Step-by-step explanation:

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...

$x^{2} + y ^ {2} + 2gx + 2fy + c = 0$

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

$(- 6) ^ {2} + 0 ^ {2 }+ 2g(- 6) + 2f(0) + c = 0$

=> 36 - 12g + c = 0 (1)

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

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

⇒ 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 :

$x^{2} + y^{2}+4x+6y - 12 = 0$..........(4)

Here, the fourth point should automatically satisfy this equation for the points to be con-cyclic.

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

We get,

(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.