Given that S1≡x2+y2+2x+3y+2=0 . . . (i) and S2≡x2+y2+2x−3y−4=0 . . . (ii) Equation of common chord is S1−S2=0 ⇒6y+6=0⇒y=−1 Putting y=−1 in Eq. (i), we get ∴x2+1+2x−3+2=0 ⇒x2+2x=0⇒x=0,−2 ∴ End points of diameter are (0,−1) and (−2,−1) Equation of circle is (x−0)(x+2)+(y+1)(y+1)=0 ⇒x2+2x+y2+2y+1=0