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

Hence the general equation of the circle is x

^{2}+ y

^{2}+ 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

x

^{2}+ y

^{2}+ 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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