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Question

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


A
0<r<11
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B
1<r<11
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C
r>11
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D
r=11
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Solution

The correct option is B $$1<r<11$$
$$x^2+y^2-16x-20y+164=r^2$$
$$A(8,10),R_1=r$$
$$(x-4)^2+(y-7)^2=36$$
$$B(4,7),R_2=6$$
$$|R_1-R_2|<AB<R_1+R_2$$
$$\Rightarrow 1<r<11$$

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