Given equation of circles ‌S1:x2+y2+2x−3y−5=0 ‌g1=1,f1=‌
−3
2
,c1=−5 ‌S2:x2+y2−3x+2y+1=0 ‌g2=‌
−3
2
,f2=1,c2=1 Let S=x2+y2+2αx+2βy+c=0 ∴‌2gg1+2ff1=c1+c2 ‌2α−3β=c−5 ‌2gg2+2ff2=c1+c2 ‌−3α+2β=c+1 ‌(1,−1),1+1+2α−2β+c=0 ‌2α−2β=−2−c By Eqs. (i) and (iii), By Eqs. (ii) and (iii), −β=2c−3‌−α=−1 ⇒β=3−2c‌⇒α=1 By Eqs. (iii), ‌⇒2−2β=−2−c ‌⇒2β=c+4 ‌⇒2(3−2c)=c+4 ‌⇒6−4c=c+4 ‌⇒5c=2⇒c=‌
