Two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2}$ and $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct points, if

2 < r < 8

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r < 2

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r = 2

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r > 2

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Solution

The correct option is A $2 < r < 8$
The center of the first circle is $(1, 3)$ and the radius is $r$ .
The center and the radius of the second circle is $(4, - 1)$ and the radius is $\sqrt{4^{2} + {(- 1)}^{2} - 8} = 3$
We know that the condition for two circles to touch externally at two distinct points is that the distance between their centers is greater than the absolute difference between their radii.
$C_{1} C_{2} < | r_{1} - r_{2} |$
$⟹ 5 > | r - 3 | ⟹ 2 < r < 8$

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