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A square is incribed in a circle x^2+y^2...

A square is incribed in a circle `x^2+y^2-6x+8y-103=0` such that its sides are parallel to co-ordinate axis then the distance of the nearest vertex to origin, is equal to (A) `13` (B) `sqrt127` (C) `sqrt41` (D) `1`

A

13

B

`sqrt(137)`

C

`6`

D

`sqrt(41)`

Text Solution

Verified by Experts

The correct Answer is:
D
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