If the radical axis of the circles x^2+y^2+2 g x+2 f y+c=0 and 2 x^2+2 y^2+3 x+8 y+2 c=0 touches the circle x^2+y^2+2 x+2 y+1=0 , then

Step-by-step solution

aligned & Let point P(a, b) & array l S₁(a, b)=S₂(a, b) a^2+b^2+2 g a+2 f b+c=a^2+b^2+ 3 2 a+4 b+c=0 a (2 g- 3 2 )+b(2 f-4)=0 array aligned this is the locus of radical axis. So, x (2 g- 3 2 )+y(2 f-4)=0 is radical axis of given circles. This touched the x^2+y^2+2 x+2 y+1=0 So, radius = 1^2+1^2-1 =1 and centre =(-1,-1) So, radius of circle = distance between centre and touching point. 1= | ( 3 2 -2 g )+(2 f-4) | ( 3 2 -2 g )^2+(2 f-4)^2 Taking square both sides, 2 ( 3 2 -2 g )(2 f-4)=0 So, aligned 3 2 -2 g & =0 or