If the two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2}$ and $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect at 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
$Circles intersect i f | r_{1} - r_{2} | < C_{1} C_{2} \Rightarrow | r - 3 | < 5 \Rightarrow 0 < r < 8 . . . (i) a n d r_{1} + r_{2} > C_{1} C_{2} \Rightarrow r + 3 > 5 \Rightarrow r > 2 . . . (i i)$
From (i) and (ii), we get
2 < r < 8

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Similar questions

(x - 1)^{2} + (y - 3)^{2} = r^{2}

x^{2} + y^{2} - 8 x + 2 y + 8 = 0

If two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2} a n d x^{2} + y^{2} - 4 x - 4 y + 4 = 0$ intersect in two distinct point then

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

(x - 1)^{2} + (y - 3)^{2} = r^{2}

x^{2} + y^{2} - 8 x + 2 y + 8 = 0

{(x - 1)}^{2} + {(y - 3)}^{2} = r^{2}

x^{2} + y^{2} - 8 x + 2 y + 8 = 0

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