Circle

The equation of a circle that passes through the intersection points of the given circles is determined as follows:
First, start with the general form that includes both circles and a parameter λ :
x2+y2−2x+2y−2+λ(x2+y2+2x−2y+1)=0
Simplifying this equation leads to:
(1+λ)x2+(1+λ)y2+(2λ−2)x+(2−2λ)y+(−2+λ)=0
To rewrite in standard circle form, factor out the coefficients:
x2+y2+2x(‌

λ−1

1+λ

)+2y(‌

1−λ

1+λ

)−‌

2−λ

1+λ

=0
The center of the circle (h,k) from this equation is:
(‌

1−λ

1+λ

,‌

λ−1

1+λ

)
Since this center lies on the line x−y+6=0, we substitute the center into the line's equation:
1−λ−(λ−1)+6(1+λ)=0
Solving:
1−λ−λ+1+6λ+6=0⇒4λ+8=0⇒λ=−2
Substitute λ=−2 back into the center, we get:
(h,k)=(−3,3)
Finally, calculate the radius of the circle:
‌ Radius ‌=√(−3)2+32−‌

2−(−2)

1+(−2)

=√(−3)2+32−‌

4

−1

Solving further:
Radius =√9+9+4=√14