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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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1 < r < 11

r > 11

0 < r < 1

r = 11

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detailed solution

Correct option is A

Given equation of circle is $x^{2} + y^{2} - 16 x - 20 y + 164 = r^{2} \Rightarrow (x - 8)^{2} + (y - 10)^{2} = r^{2}$

$C_{1} = (8, 10)$ and $R_{1} = r$

for the second circle $C_{2}$ =(4,7), $R_{2}$ =6

now, $C_{1} C_{2} = \sqrt{(8 - 4)^{2} + (10 - 7)^{2}} = \sqrt{25} = 5$

Since the circles intersect at two distinct points So,

$R_{1} + R_{2} > C_{1} C_{2} > |R_{1} - R_{2}|$

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

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If the circles x2+y2−16x−20y+164=r2 and (x−4)2+(y−7)2=36 intersect at two distinct points, then