lf the 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 then

r < 2

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r = 4

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r > 2

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r = 2

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Solution

The correct option is B $r > 2$
$S_{1} : (x - 1)^{2} + (y - 3)^{2} = r^{2}$
$C_{1} \equiv (1, 3)$ and $r_{1} = r a d i u s = e$
$S_{2} x^{2} + y^{2} - 8 x + 2 y + 8 = 0$
$C_{2} \equiv (4, - 1)$ and $r_{2} = r a d i u s = 3$
Given $S_{1}$ & $S_{2}$ intersect in two distinct points than,
$r_{1} + r_{2} > C_{1} C_{2}$
$∴ r + 3 > \sqrt{4^{2} + 3^{2}}$
$r > 5 - 3$
$r > 2$

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