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

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

Apply $r_{1} - r_{2}$ < d < $r_{1} + r_{2}$

When d is then distance between centres

Suggest Corrections

Similar questions

(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

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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