Answer:
[tex]x^{2} +y^{2} +(-8)\times x+ (-6)\times y = 0[/tex]
Step-by-step explanation:
General equation of circle is [tex]x^{2} +y^{2} +2\times g \times x+2\times f\times y+c = 0[/tex] ---------------------(1)
This circle passes through the given three points, (1,7) , (8,6) and (7,-1).
Substitute (1,7) in (1),
1 + [tex]7^{2} +2\times g\times 1 + 2\times f\times 7 +c[/tex] =0
2g + 14f + c = -50 -------------(2)
Substitute (8,6) in (1),
[tex]8^{2} + 6^{2} +2\times g \times 8+2\times f\times 6 +c=0[/tex]
16g + 12f + c = -100 -----------(3)
Substitute (7,-1) in (1),
[tex]7^{2} + (-1)^{2} +2\times g \times 7+2\times f\times (-1) +c=0[/tex]
14g - 2f + c = -50 ---------------(4)
Solving (2),(3) amd (4) simultaneously,
g = -4
f = -3
c = 0
Equation of circle is
[tex]x^{2} +y^{2} +(-8)\times x+ (-6)\times y = 0[/tex]