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 $\lambda$ :$x^2+y^2-2x+2y-2+\lambda (x^2+y^2+2x-2y+1)=0$Simplifying this equation leads to:$(1+\lambda)x^2+(1+\lambda)y^2+(2\lambda-2)x+(2-2\lambda)y+(-2+\lambda)=0$To rewrite in standard circle form, factor out the coefficients:$x^2+y^2+2x\left(\frac{\lambda-1}{1+\lambda}\right)+2y\left(\frac{1-\lambda}{1+\lambda}\right)-\frac{2-\lambda}{1+\lambda}=0$The center of the circle $(h,k)$ from this equation is:$\left(\frac{1-\lambda}{1+\lambda},\frac{\lambda-1}{1+\lambda}\right)$Since this center lies on the line $x-y+6=0$, we substitute the center into the line's equation:$1-\lambda-(\lambda-1)+6(1+\lambda)=0$Solving:$1-\lambda-\lambda+1+6\lambda+6=0 \Rightarrow 4\lambda+8=0 \Rightarrow \lambda=-2$Substitute $\lambda=-2$ back into the center, we get:$(h,k)=(-3,3)$Finally, calculate the radius of the circle:$\text{ Radius }=\sqrt{(-3)^2+3^2-\frac{2-(-2)}{1+(-2)}}=\sqrt{(-3)^2+3^2-\frac{4}{-1}}$Solving further:Radius $=\sqrt{9+9+4}=\sqrt{14}$