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The shortest distance between the circles \mathrm{x^{2}+y^{2}=1} and \mathrm{x^{2}+y^{2}-10 x-10 y+41=0} is

Option: 1

\sqrt{41}-1


Option: 2

0


Option: 3

\sqrt{41}


Option: 4

5 \sqrt{2}-4


Answers (1)

best_answer

The distance between the centres of the circles is 5 \sqrt{2}.

Sum of the radii =1+3=4.

Hence, shortest distance =5 \sqrt{2}-4.

Hence (D) is the correct answer.

 

Posted by

HARSH KANKARIA

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