The equation of a circle passing through the intersection of two circles ‌x2+y2+2x+4y+1=0‌ and ‌ ‌x2+y2−2x−4y−4=0‌ is ‌ ‌(x2+y2+2x+4y+1) ‌+λ(x2+y2−2x−4y−4)=0,‌ where ‌λ is a parameter. ⇒(1+λ)x2‌+(1+λ)y2+(2−2λ)+(4−4λ)y+1−4λ=0 ‌⇒x2+y2+‌
2(1−λ)
(1+λ)
x+‌
4(1−λ)
(1+λ)
y+‌
1−4λ
1+λ
=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(i) Applying the orthogonal condition, 2(‌
1−λ
1+λ
)×0‌+2×‌
2(1−λ)
(1+λ)
×0 ‌=‌
1−4λ
1+λ
−6 ⇒‌‌‌
1−4λ
1+λ
‌=6 ⇒‌‌6+6λ‌=1−4λ⇒λ=‌
−1
2
Substitutingλ=‌
−1
2
in Eq. (i) and simplifying we get x2+y2+6x+12y+6=0 ∴ Radius of this circle =√32+62−6=√39