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If the circles $x^{2} + y^{2} - 16 x - 20 y + 164 = r^{2}$ and ${(x - 4)}^{2} + {(y - 7)}^{2} = 36$ intersect at two distinct points, then

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$r > 11$

$r = 11$

$1 < r < 11$

$0 < r < 1$

answer is B.

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Detailed Solution

$c_{1} = (8, 10), r_{1} = r, c_{2} = (4, 7), r_{2} = 6$

Two circles intersect each other

$| r_{1} - r_{2} | < c_{1} c_{2} < r_{1} + r_{2}$

$\Rightarrow | r - 6 | < 5 < r + 6 \Rightarrow r > 1, r < 11 \Rightarrow 1 < r < 11$

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