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 square is incribed in a circle `x^2+y^2-6x+8y-103=0` such that its s

1.	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`
Answer» equation of circle `x^2+y^2-6x+8y-103=0` centre(3,-4) `r^2=8sqrt2` `r=8sqrt2` AP=PC=r=`8sqrt2` AC=`16sqrt2` side=16 OA=`sqrt((-5)^2+(-12)^2)=13` OB=`sqrt(11^2+12^2)=sqrt265` OC=`sqrt(11^2+12^2)=sqrt137` OD=`sqrt(5^2+4^2)=sqrt41`

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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`

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