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A square is inscribed in the circle `x^(2)+y^(2)-6x+8y-103=0` with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is

A

`sqrt(41)`

B

13

C

`sqrt(137)`

D

6

Text Solution

Verified by Experts

The correct Answer is:
A

Centre of given circle (3,-4) and Radius =`8sqrt2`
Vertices of inscribed square will be `(3 pm 8 , -4 pm 8) " " { because OP=PC=(8sqrt2)/sqrt2=8}`

So, vertex nearest to origin will be (-5,4)
Hence required distance =`sqrt41`
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