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(A) Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$. Since it cuts the given circles orthogonally,we use the condition $2g_1g_2 + 2f_

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$\frac{\sqrt{193}}{4 \sqrt{2}}$

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(A) Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$. Since it cuts the given circles orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$. For $x^2+y^2-4x-4y+7=0$: $2g(-2) + 2f(-2) = c+7$ $\Rightarrow -4g-4f-c=7$ $\Rightarrow 4g+4f+c=-7$ $(i)$. For $x^2+y^2+4x-4y+6=0$: $2g(2) + 2f(-2) = c+6 \Rightarrow 4g-4f-c=6$ $(ii)$. For $x^2+y^2+4x+4y+5=0$: $2g(2) + 2f(2) = c+5 \Rightarrow 4g+4f-c=5$ $(iii)$. Subtracting $(iii)$ from $(i)$: $(4g+4f+c) - (4g+4f-c) = -7-5$ $\Rightarrow 2c = -12$ $\Rightarrow c = -6$. Substituting $c=-6$ into $(i)$ and $(ii)$: $4g+4f = -1$ and $4g-4f = 0$. Adding these: $8g = -1 \Rightarrow g = -\frac{1}{8}$. Subtracting: $8f = -1 \Rightarrow f = -\frac{1}{8}$. The radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-\frac{1}{8})^2 + (-\frac{1}{8})^2 - (-6)} = \sqrt{\frac{1}{64} + \frac{1}{64} + 6} = \sqrt{\frac{2}{64} + 6} = \sqrt{\frac{1}{32} + 6} = \sqrt{\frac{193}{32}} = \frac{\sqrt{193}}{4\sqrt{2}}$.

The radius of the circle which cuts the circles $x^2+y^2-4x-4y+7=0$,$x^2+y^2+4x-4y+6=0$,and $x^2+y^2+4x+4y+5=0$ orthogonally is

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