The correct option is A x2+y2−17x−19y+50=0
Let the equation of side AB,BC and CA of △ABC are respectively
x+y=6,
2x+y=4
and x+2y=5
The coordinates of A,B and C are (7,−1),(−2,8) and (1,2) respectively.
Let the equation of the circumcircle of △ABC be
x2+y2+2gx+2fy+c=0 …(i)
It passes through A(7,−1),B(−2,8) and C(1,2)
Therefore,
50+14g−2f+c=0
68−4g+16f+c=0
and 5+2g+4f+6c=0
On solving these equations, we get
g=−172,f=−192 and c=50
On substituting the value of g,f and c in Equation (i), the equation of the required circumcircle is
x2+y2−17x−19y+50=0
Aliternative
Consider the equation
(x+y−6)(2x+y−4)+λ(2x+y−4)(x+2y−5)+μ(x+2y−5)(x+y−6)=0
where λ and μ are constants.
This equation represents a curve passing through the vertices of the triangle formed by the given lines.
We have to determine the values of λ and μ so that the curve given in Equation (i) is a circle.
The curve given in Equation (i) will be a circle, if
coefficient of x2 = coefficient of y2
and coefficient of xy=0
⇒2+2λ+μ=1+2λ+2μ3+5λ+3μ=0
⇒μ=1 and 3μ+5λ+3=0
⇒μ=1 and λ=−65
On substituting the values of λ and μ in Eq. (i), we get
(x+y−6)(2x+y−4)−(65)(2x+y−4)(x+2y−5)+(x+2y−5)(x+y−6)=0
⇒x2+y2−17x−19y+50=0