The correct option is A x2+y2−17x−19y+50=0
Let the sides of △ABC be
AB:x+y=6 ⋯(i)BC:2x+y=4 ⋯(ii)AC:x+2y=5 ⋯(iii)
The coordinates of A,B and C are
A=(7,−1)B=(−2,8)C=(1,2)
Let the equation of the circumcircle of △ABC be
x2+y2+2gx+2fy+c=0
It passes through A(7,−1),B(−2,8) and C(1,2), so
50+14g−2f+c=0 ⋯(iv)68−4g+16f+c=0 ⋯(v)5+2g+4f+c=0 ⋯(vi)
On solving these equations, we get
g=−172,f=−192 and c=50
Hence, the required equation of the circumcircle is
x2+y2−17x−19y+50=0