The correct option is A 4x2+4y2+30x−13y−25=0
the equation of any curve through the points of intersection of the circles (1) and (2) will be
(x2+y2+13x−3y)+k(2x2+2y2+4x−7y−25)=0
It is given that equation of circles passes through the point (1,1)
So,
x=1 and y=1
Substitute these value in an above equation, we get
(1+1+13−3)+k(2+2+4−7−25)=0
12−24k=0
k=12
Now, Subsitute the value of k in an first equation, we get
(x2+y2+13x−3y)+12(2x2+2y2+4x−7y−25)=0
2x2+2y2+15x−13y2−252=0
4x2+4y2+30x−13y−25=0