If a circle has its centre on the line x-y-1= | vedclass

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(B) The family of circles passing through the points of intersection of two circles $S_1: x^2+y^2+2x-2y-2=0$ and $S_2: x^2+y^2-2x+2y-7=0$ is given by

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$\left(\frac{1}{2}, \frac{-1}{2}\right)$

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(B) The family of circles passing through the points of intersection of two circles $S_1: x^2+y^2+2x-2y-2=0$ and $S_2: x^2+y^2-2x+2y-7=0$ is given by $S_1 + \lambda S_2 = 0$. $(x^2+y^2+2x-2y-2) + \lambda(x^2+y^2-2x+2y-7) = 0$ $(1+\lambda)x^2 + (1+\lambda)y^2 + 2(1-\lambda)x + 2(\lambda-1)y - (2+7\lambda) = 0$ Dividing by $(1+\lambda)$,we get the equation of the circle: $x^2 + y^2 + 2\left(\frac{1-\lambda}{1+\lambda}\right)x + 2\left(\frac{\lambda-1}{1+\lambda}\right)y - \frac{2+7\lambda}{1+\lambda} = 0$ The centre of this circle is $\left(-\frac{1-\lambda}{1+\lambda}, -\frac{\lambda-1}{1+\lambda}\right) = \left(\frac{\lambda-1}{\lambda+1}, \frac{1-\lambda}{\lambda+1}\right)$. Since the centre lies on the line $x-y-1=0$,we substitute the coordinates: $\frac{\lambda-1}{\lambda+1} - \frac{1-\lambda}{\lambda+1} - 1 = 0$ $\frac{\lambda-1 - 1 + \lambda - \lambda - 1}{\lambda+1} = 0$ $\lambda - 3 = 0 \Rightarrow \lambda = 3$. Substituting $\lambda = 3$ into the centre coordinates: $x = \frac{3-1}{3+1} = \frac{2}{4} = \frac{1}{2}$ $y = \frac{1-3}{3+1} = \frac{-2}{4} = -\frac{1}{2}$ Thus,the centre is $\left(\frac{1}{2}, -\frac{1}{2}\right)$.

If a circle has its centre on the line $x-y-1=0$ and passes through the points of intersection of the two circles $x^2+y^2+2x-2y-2=0$ and $x^2+y^2-2x+2y-7=0$,then the centre of that circle is

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