If the circle x2+y2+2x+3y+1=0 cuts x2+y2+4x+3y+2=0 in A and B, then the equation of the circle on AB as diameter, is

The equation of the common chord AB of the two circles is       2x+1=0                            [Using S1-S2=0]The equation of the required circle is     x2+y2+2x+3y+1+λ(2x+1)=0                                                 [Using : S1+λ(S2-S1)=0]⇒ x2+y2+2x(λ+1)+3y+λ+1=0.Since AB is a diameter of this circle. Therefore, centre of this circle lies on AB.∴ −2λ−2+1=0⇒λ=−1/2So, the equation of the required circle is    x2+y2+x+3y+1/2=0or, 2x2+2y2+2x+6y+1=0