The equation of the circles are S1≡x2+y2+2x+3y+1=0 and S2≡x2+y2+4x+3y+2=0 Since, the circles cuts each other at A and B then equation of AB is ‌S1−S2=0 ⇒(x2+y2+2x+3y+1) ⇒−(x2+y2+4x+3y+2)=0 ⇒−2x−1=0 ⇒x=‌
−1
2
Putting the value x=−‌
1
2
in S2, we get ‌‌‌‌
1
4
+y2−2+3y+2=0 ⇒‌‌4y2+12y+1=0 ⇒y=‌
−12±√144−16
8
⇒y=‌
−12±√128
8
=‌
−12±8√2
8
⇒y=‌
−3
2
±√2 So intersection points are A(−‌
1
2
,−‌
3
2
+√2) and (−‌
1
2
,−‌
3
2
−√2). Then equation of circle with diameter AB is ‌(x+‌