Consider the given circles, S1≡x2+y2+2x+2y+1=0 And, S2≡x2+y2+4x+6y+4=0 The equation of common chord of these two circles, s1−s2=0 −2x−4y−3=0 2x+4y+3=0 The equation of the required circle is, S1+λ(s1−s2)=0 (x2+y2+2x+2y+1)+λ(2x+4y+3)=0 x2+y2+2x(1+λ)+2y(2λ+1)+3λ+1=0 Now, 2x+4y+3=0 is the diameter of the circle. So, 2(−(1+λ))+4(−(2λ+1))+3=0 −2−2λ−8λ−4+3=0 10λ=−3 λ=−
3
10
Therefore, the equation of the required circle is, x2+y2+2x(1−