# 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 :Option 1)  $\sqrt{41}$Option 2)13Option 3)  $\sqrt{137}$Option 4)  6

General form of a circle -

$x^{2}+y^{2}+2gx+2fy+c= 0$

- wherein

centre = $\left ( -g,-f \right )$

radius = $\sqrt{g^{2}+f^{2}-c}$

Radius of circle, $R=\sqrt{r^{2}+f^{2}-c}$

$R=\sqrt{9+16+103}$

$=\sqrt{128}=8\sqrt{2}$

circle equation can be written as -

$(x-3)^{2}+(y+4)^{2}=(8\sqrt{2})^{2}$

$OC=\sqrt{11^{2}+42}=\sqrt{137}$

$OO=\sqrt{25+16}=\sqrt{41}$

Nearest vertex from origin is $\sqrt{41}$

Option 1)

$\sqrt{41}$

Option 2)

13

Option 3)

$\sqrt{137}$

Option 4)

6

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