If the cicles $(x - 1)^{2} + (y - 3)^{2} = r^{2} a n d x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct points, then

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

If d is distance between the centers of two circles os radii $r_{1} a n d r_{2}$ ,then they intersect in two distinct

points if $| r_{1} - r_{2} | < d < r_{1} + r_{2}$

Center of circles are $C_{1} a n d C_{2}$

$C_{1} = (1, 3) a n d C_{2} = (4, - 1)$

$C_{1} C_{2} = \sqrt{(1 - 4)^{2} + (3 + 1)^{1}} = \sqrt{9 + 16} = 5$

$r_{1} = r a n d r_{2} = \sqrt{g^{2} + f^{2} - c} = \sqrt{16 + 1 - 8} = 3$

Here radii of two circles are r and 3 and distance between the centers is 5.

Thus, $| r - 3 | < 5 < r + 3 \Rightarrow - 2 < r < 8 a n d r > 2 \Rightarrow 2 < r < 8$

Hence the correct answer is (a)

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