Question

Show that the points (1,-6)(5,2)(7,0)(-1,-4)are concyclic and find equation of circle​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The points (1,-6)(5,2)(7,0)(-1,-4) are concyclic and the equation of the circle is $x^{2} + y^{2} - 6x + 4y -7 =0$

Step-by-step explanation:

Given: The given points are  the points (1,-6)(5,2)(7,0)(-1,-4)

To find: The points are concyclic and to find the equation of the circle

Solution:

The points are  (1,-6), (5,2), (7,0), (-1,-4) are concyclic

Let equation of circle passing through the these points

$x^{2} + y^{2} + 2gx +2fy +c = 0$           (1)

If all points lie on circle then satisfy equation (1)

(1, -6)

⇒ $x^{2} + y^{2} + 2gx +2fy +c = 0$

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

⇒ 1 +36 +2g -12f + c =0

⇒ 2g - 12f +c +  37 = 0        (2)

(5,2)

⇒ $x^{2} + y^{2} + 2gx +2fy +c = 0$

⇒ $(5)^{2} + (2)^{2} + 2g(5) +2f(2) +c =0$

⇒ 25+4 +10g +4f + c =0

⇒10g + 4f + c +29 =0           (3)

(7,0)

⇒ $x^{2} + y^{2} + 2gx +2fy +c = 0$

⇒ $(7)^{2} + (0)^{2} + 2g(7) +2f(0) +c =0$

⇒ 49+0+14g +0+ c =0

⇒14g + c +49 =0  

c = -49 - 14g         (4)

(-1,-4)

⇒ $x^{2} + y^{2} + 2gx +2fy +c = 0$

⇒ $(-1)^{2} + (-4)^{2} + 2g(-1) +2f(-4) +c =0$

⇒ 1+16 - 2g - 8f + c =0

⇒-2g -8f +c + 17 =0

Sub eq 4  in eq 2

2g - 12f +c +  37 = 0  

c = -49 - 14g

2g - 12f -49 - 14g+  37 = 0

⇒-12g -12f -12 = 0

⇒ - g - f - 1 = 0    

g + f+ 1 = 0        (5)

Sub eq 4  in eq 3

10g + 4f + c +29 =0  

⇒ 10g + 4f -49 - 14g +29 = 0

⇒ -4g + 4f -20 = 0

⇒ -g +f - 5 = 0

g -f +5 =0        (6)

Add eq 5 and eq 6

g + f+ 1 = 0  

g -f +5 =0

----------------

2g  + 6 =0

2g = -6

g = -3

Sub g = 3 in eq 5

g + f+ 1 = 0

-3 + f +1 = 0

-2 +f = 0

f = 2

Sub the value of f and g in eq 4

c = -49 - 14g

c = -49 - 14 (-3)

c = -49 + 42

c = -7

Sub the value of g , f and c in eq 1

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

$x^{2} + y^{2} + 2(-3)x +2(2)y -7 = 0$

$x^{2} + y^{2} - 6x + 4y -7 =0$

Sub the points in above equation

$x^{2} + y^{2} - 6x + 4y -7 =0$

(-1 , -4)

$(-1)^{2} + (-4)^{2} - 6(-1) +4 (-4) - 7 =0$

1 + 16 + 6 -16 -7 = 0

7 + 16 - 16 - 7 = 0

0 = 0

As the result  the points (1,-6)(5,2)(7,0)(-1,-4)are concyclic

Final answer:

The points (1,-6)(5,2)(7,0)(-1,-4)are concyclic and the equation of the circle is $x^{2} + y^{2} - 6x + 4y -7 =0$

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