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(D) The equation of the family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + kS_2 = 0$.$(x^2+y^2-2

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$\sqrt{14}$

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(D) The equation of the family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + kS_2 = 0$.$(x^2+y^2-2x+2y-2) + k(x^2+y^2+2x-2y+1) = 0$$(1+k)x^2 + (1+k)y^2 + (2k-2)x + (2-2k)y + (k-2) = 0$Dividing by $(1+k)$,we get $x^2 + y^2 + \frac{2(k-1)}{k+1}x + \frac{2(1-k)}{k+1}y + \frac{k-2}{k+1} = 0$.The centre of this circle is $\left(-\frac{k-1}{k+1}, -\frac{1-k}{k+1}\right) = \left(\frac{1-k}{k+1}, \frac{k-1}{k+1}\right)$.Since the centre lies on the line $x-y+6=0$,we substitute the coordinates:$\frac{1-k}{k+1} - \frac{k-1}{k+1} + 6 = 0$$\frac{1-k-k+1}{k+1} = -6 \Rightarrow 2-2k = -6k-6$$4k = -8 \Rightarrow k = -2$.Substituting $k=-2$ into the equation: $x^2+y^2 + \frac{2(-3)}{-1}x + \frac{2(3)}{-1}y + \frac{-4}{-1} = 0 \Rightarrow x^2+y^2+6x-6y+4=0$.The radius $r = \sqrt{g^2+f^2-c} = \sqrt{3^2+(-3)^2-4} = \sqrt{9+9-4} = \sqrt{14}$.

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

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