Let $x^{2} + y^{2} - 2 x - 6 y + 9 = 0$ and $x^{2} + y^{2} + 6 x - 2 y + 1 = 0$ be the equation of two circles. Then

circles touch each other internally

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circles touch each other externally

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circles neither touch nor intersect each other.

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circles are concentric.

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Solution

The correct option is C circles neither touch nor intersect each other.
Let $S_{1} : x^{2} + y^{2} - 2 x - 6 y + 9 = 0$
Centre $C_{1} \equiv (1, 3)$
and radius $r_{1} = \sqrt{(- 1)^{2} + (- 3)^{3} - 9} = 1$

$S_{2} : x^{2} + y^{2} + 6 x - 2 y + 1 = 0$
Centre $C_{2} \equiv (- 3, 1)$
and radius $r_{2} = \sqrt{3^{2} + (- 1)^{2} - 1} = 3$

Now, $C_{1} C_{2} = \sqrt{(1 + 3)^{2} + (3 - 1)^{2}} = \sqrt{16 + 4}$
$\Rightarrow C_{1} C_{2} = \sqrt{20} = 2 \sqrt{5}$
and $r_{1} + r_{2} = 1 + 3 = 4$
Since $C_{1} C_{2} > r_{1} + r_{2},$
hence circles neither touch nor intersect each other.

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