Circle

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