A Circle S passes through the points of intersection of the circles x2+y2−2x+2y−2=0 and x2+y2+2x−2y+1=0. If the centre of this circle S lies on the line x−y+6=0, then the radius of the circle S is
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−‌