The correct option is A x2+y2−17x−19y+50=0
Given lines
L1:x+y−6=0L2:2x+y−4=0L3:x+2y−5=0
Consider the equation
L1L2+λL2L3+μL3L1=0
⇒(x+y−6)(2x+y−4) +λ(2x+y−4)(x+2y−5) +μ(x+2y−5)(x+y−6)=0 ⋯(i)
Where λ and μ are constants.
This is a circle, when
coefficient of x2= coefficient of y2
⇒2+2λ+μ=1+2λ+2μ⇒μ=1
Coefficient of xy=0
⇒3+5λ+3μ=0⇒λ=−65
On substituting the values of λ and μ in equation (i), we get
L1L2−65L2L3+L3L1=0⇒5L1[L2+L3]=6L2L3⇒5(x+y−6)3(x+y−3)=6(2x+y−4)(x+2y−5)⇒5[x2+y2+2xy−9x−9y+18] =2[2x2+2y2+5xy−14x−13y+20]⇒x2+y2−17x−19y+50=0
Alternate Solution:
Let the equation of side be
AB:x+y=6BC:2x+y=4AC:x+2y=5
Finding the point of intersection, we get the coordinates of the vertices as
A=(7,−1), B=(−2,8), C=(1,2)
Let the equation of the circumcircle of △ABC be
x2+y2+2gx+2fy+c=0 ⋯(ii)
It passes through A(7,−1),B(−2,8) and C(1,2), so
50+14g−2f+c=068−4g+16f+c=05+2g+4f+6c=0
On solving these equations, we get
g=−172,f=−192, c=50
Hence, the equation of the circumcircle is
x2+y2−17x−19y+50=0