Given, s≡x2+y2+2gx+2fy+c=0 Ands′≡x2+y2−6x+6y+2=0 ‌⇒‌‌(x2−6x)+(y2+6y)=−2 ‌⇒‌‌(x2−6x+9)+(y2+6y+9) ‌=−2+9+9 ‌⇒‌‌(x−3)2+(y+3)2=16 Thus, the center of circleS′is(3,−3)and radius,r′=4 Now,S≡x2+y2+2gx+2fy+c=0 ⇒‌‌(x+g)2+(y+f)2=g2+f2−c So, the center of circleSis(−g,−f)and radius, r=√g2+f2−c Since, radius of circle S is 1 . ∴‌√g2+f2−c‌=1 ⇒‌g2+f2−c‌=1 Also, two circle touch externally, So, distance between (−g,−f) and ‌(3,−3)=r+r′ ‌⇒‌‌√(−g−3)2+(−f+3)2=4+1 ‌⇒‌‌(g+3)2+(f−3)2=25 ‌⇒‌‌g2+6g+9+f2−6f+9=25 ‌⇒‌‌g2+f2+6g−6f+18−25=0 ‌⇒‌‌g2+f2+6g−6f=7‌‌‌⋅⋅⋅⋅⋅⋅⋅(i) Since, point(−1,−3)lies on both circles, it satisfies both circle. ∴ For circles ‌(−1)2+(−3)2+2g(−1)+2f(−3)+c=0 ‌⇒‌‌1+9−2g−6f+c=0 ‌⇒‌‌−2g−6f+c=−10‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii) For circle S′, ‌(−1)2+(−3)2−6(−1)+6(−3)+2=0 ‌⇒‌‌1+9+6−18+2=0 ‌⇒‌‌0=0 Now, the slope of line joining (−g,−f) and (−1,−3) is equal to slope of line joining (−1,−3) and (3,−3) ‌(−f+3)‌=(‌
−3+3
3+1
)(−g+1) ⇒‌(−f+3)‌=0 ⇒‌f‌=3 Put f=3 into Eq. (i), we get ‌g2+9+6g−6(3) ‌⇒g2+9+6g−18‌=7 ‌⇒g2+6g−16‌=0 ‌⇒(g+8)(g−2)‌=0 ‌⇒g‌=−8‌ or ‌g=2‌ (not possible) ‌ Now, put g=2,f=3 into Eq. (ii), we get −2(2)−6(3)+c‌=−10 ⇒‌‌c‌c=−10+4+18=12 Now, g+f+c=2+3+12=17
