The equation of four circles are x- 4 2 - y- 4 2 - 4 2- then the radius of the smallest circle touching all the four circle is

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Question

The equation of four circles are ${(x \pm 4)}^{2} + {(y \pm 4)}^{2} = 4^{2}$ , then the radius of the smallest circle touching all the four circle is

4 (\sqrt{2} + 1)

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4 (\sqrt{2} - 1)

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2 (\sqrt{2} - 1)

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None of these

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Solution

The correct option is B $4 (\sqrt{2} - 1)$
Clearly, from the figure, the radius of the smallest circle touching the given circle is
$= \sqrt{4^{2} + 4^{2}} - 4 \Rightarrow 4 \sqrt{2} - 4$

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The equation of four circles are (x±4)2+(y±4)2=42, then the radius of the smallest circle touching all the four circle is

The correct option is B 4(√2−1)Clearly, from the figure, the radius of the smallest circle touching the given circle is =√42+42−4⇒4√2−4

The equation of four circles are ${(x \pm 4)}^{2} + {(y \pm 4)}^{2} = 4^{2}$ , then the radius of the smallest circle touching all the four circle is

The correct option is B $4 (\sqrt{2} - 1)$
Clearly, from the figure, the radius of the smallest circle touching the given circle is
$= \sqrt{4^{2} + 4^{2}} - 4 \Rightarrow 4 \sqrt{2} - 4$