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Minimum distance between y^2-4x - 8y +40...

Minimum distance between` y^2-4x - 8y +40 =0` and `x^2 - 8x-4y +40=0` is

A

0

B

`sqrt(3)`

C

`2sqrt(2)`

D

`sqrt(2)`

Text Solution

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The correct Answer is:
D
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