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 point then

0 < r < 11

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1 < r < 11

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r > 11

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r = 11

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Solution

The correct option is B $1 < r < 11$
$x^{2} + y^{2} - 16 x - 20 y + 164 = r^{2}$
$A (8, 10), R_{1} = r$
$(x - 4)^{2} + (y - 7)^{2} = 36$
$B (4, 7), R_{2} = 6$
$| R_{1} - R_{2} | < A B < R_{1} + R_{2}$
$\Rightarrow 1 < r < 11$

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