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JEE Advance - Mathematics (1989 - No. 20)

If the two circles $${(x - 1)^2} + {(y - 3)^2} = {r^2}$$ and $${x^2} + {y^2} - 8x + 2y + 8 = 0$$ intersect in two distinct points, then
2 < r < 8
r < 2
r = 2
r > 2

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