The given two spheres $x^{2} + y^{2} + z^{2} = 64$ , $x^{2} + y^{2} + z^{2} - 12 x + 4 y - 6 z + 48 = 0$

touch internally

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touch externally

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intersect each other

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do not intersect, do not touch

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Solution

The correct option is A touch internally
From the given $2$ spheres, ....... $∵ c_{1} = (0, 0, 0), r_{1} = 8$
$c_{2} = (6, - 2, 3), r_{2} = \sqrt{36 + 4 + 9 - 48} = 1$
As $c_{1} c_{2} = \sqrt{36 + 4 + 9} = 7$ & $| r_{1} - r_{2} | = 7$
$∵ c_{1} c_{2} = | r_{1} - r_{2} | = 7$
$\Rightarrow$ Spheres touches internally

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Similar questions

x^{2} + y^{2} + z^{2} = 64

x^{2} + y^{2} + z^{2} - 12 x + 4 y - 6 z + 48 = 0

x^{2} + y^{2} + z^{2} - 2 x + 4 y + 6 z + 1 = 0

The two circles $x^{2} + y^{2} - 4 y = 0$ and $x^{2} + y^{2} - 8 y = 0$

x^{2} + y^{2} + z^{2} - 2 x + 4 y - 6 z + 7 = 0

12 x + 4 y + 3 z = 327

x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z = 155

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