A square is inscribed in the circle $x^{2}+y^{2}-6 x+8 y-103=0$ with its sides parallel to the coordinate…
A square is inscribed in the circle $x^{2}+y^{2}-6 x+8 y-103=0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is:
6
$\sqrt{137}$
$\sqrt{41}$
13
Solution
The equation of circle is, $x^{2}+y^{2}-6 x+8 y-103=0$
$\Rightarrow(x-3)^{2}+(y+4)^{2}=(8 \sqrt{2})^{2}$
$C(3,-4), r=8 \sqrt{2}$
$\Rightarrow$ Length of side of square $=\sqrt{2} r=16$
$\Rightarrow P(-5,4), Q(-5,-12)$
$R(11,-12), S(11,4)$
$\Rightarrow$ Required distance $=O P$
$=\sqrt{(-5-0)^{2}+(-4-0)^{2}}=\sqrt{25+16}=\sqrt{41}$
