Find the equation of family of circles through the intersection of x2 + y2 − 6x + 2y + 4 = 0 and x2 + y2 + 2x − 4y − 6 = 0 whose center lies on y = x.
x2+y2−107x−107y−127=0
Given circles
S1=0 x2+y2−6x+2y+4=0 - - - - - - -(1)
S2=0 x2+y2−2x−4y−6=0 - - - - - - -(2)
The equation of family of circles passing through the points of intersection of two circles S1 =0 & S2=0 is S1+λS2=0
(x2+y2−6x+2y+4)+λ(x2+y2+2x−4y−6)=0 - - - - - - - (3)
(1+λ)x2+(1+λ)y2+(2λ−6)x+(2−4λ)y+4−6λ=0
Dividing both sides by 1+λ
x2+y2+2(λ−3)1+λx+2(1−2λ)1+λy+4−6λ1+λ=0
Coordinates of center of circle is (-g,-f)
(3−λ1+λ,2λ−11+λ)
Since, center lies on the line y=x
3−λ1+λ=2λ−11+λ
{λ≠−1}
λ=43
Equation of family of circles passing through the point of intersection of circle is
(x2+y2−6x+2y+4)+43(x2+y2+2x−4y−6)=0
73x2+73y2−103x−103y−4=0
x2+y2−107x−107y−127=0
Option A is correct