If the line x+y=2 cuts the circle x^2+y^2+2x- | vedclass

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(C) The equation of the family of circles passing through the intersection of the circle $S_1: x^2+y^2+2x-4y+4=0$ and the line $L: x+y-2=0$ is given b

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$5$

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(C) The equation of the family of circles passing through the intersection of the circle $S_1: x^2+y^2+2x-4y+4=0$ and the line $L: x+y-2=0$ is given by $S_1 + \lambda L = 0$.$x^2+y^2+2x-4y+4 + \lambda(x+y-2) = 0$$x^2+y^2+(2+\lambda)x + (\lambda-4)y + (4-2\lambda) = 0$.This circle is orthogonal to $S_2: x^2+y^2-2x-4y-4=0$.The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here,$g_1 = \frac{2+\lambda}{2}$,$f_1 = \frac{\lambda-4}{2}$,$c_1 = 4-2\lambda$ and $g_2 = -1$,$f_2 = -2$,$c_2 = -4$.Substituting these values: $2(\frac{2+\lambda}{2})(-1) + 2(\frac{\lambda-4}{2})(-2) = 4-2\lambda - 4$.$-(2+\lambda) - 2(\lambda-4) = -2\lambda$.$-2-\lambda - 2\lambda + 8 = -2\lambda$.$6 - 3\lambda = -2\lambda \implies \lambda = 6$.Substituting $\lambda = 6$ into the circle equation: $x^2+y^2+8x+2y-8=0$.The radius $r = \sqrt{g^2+f^2-c} = \sqrt{4^2+1^2-(-8)} = \sqrt{16+1+8} = \sqrt{25} = 5$.

If the line $x+y=2$ cuts the circle $x^2+y^2+2x-4y+4=0$ at two points $A$ and $B$,then the radius of the circle passing through $A$ and $B$ and orthogonal to $x^2+y^2-2x-4y-4=0$ is

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