Prove that (-1,6), (5,2), (7,0), and (-1,-4) are concyclic.

For 4 points A (-1,6), B(5,2), C(7,0) and D (-1,-4) to be concyclic they should lie on the circle.
Hence the general equation of the circle is x2 + y2 + 2gx + 2fy + c = 0
If A lies on the circle then it should satisfy the general eqaution:
(-1)2+(6)2+2g(-1)+2f(6)+c=0
=> 1+36-2g+12f+c=0
=>-2g+12f+c=-37..................(1)
If B lies on the circle then it should satisfy the general equation:
(5)2+(2)2+2g(5)+2f(2)+c=0
=> 25+4+10g+4f+c=0
=> 10g + 4f + c=-29..................(2)
If C lies on the circle then it should satisfy the general equation:
(7)2+(0)2+2g(7)+2f(0)+c=0
=> 49+14g+c=0
=> 14g+c=-49.....................(3)
Solving (1)and (2) we have
12g-8f=8.............................(4)
Solving (2) and (3) we have
4g-4f=-20..................................(5)
Now solving (4) and (5) we get
f=17
and putting values we have
f=17,g=12,c=-217
So the equation becomes
x2 + y2 + 24x + 34y -217= 0

Put D (-1,-4), we are not getting 0, so the point are not concyclic, please check your question.

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