A square is inscribed in the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}-93=0$ with its sides parallel to the coordinate axes. Then  coordinates of its vertices are

$\text{ Equation of circle is }(\mathrm{x}-1{)}^2+(\mathrm{y}+2{)}^2=98\text{. }$
Square is inscribed in the given circle such that its sides are parallel to coordinate axes
$\text{ So, slope of }\mathrm{PR}\text{ is }1\text{ and that of }\mathrm{SQ}\text{ is }-1\text{. }$
 

Thus, vertices of square are
$\left(1\pm 7\sqrt{2}\cos {45}^{\circ },-2\pm 7\sqrt{2}\sin {45}^{\circ }\right)\equiv (8,5)\text{ and }(-6,-9)\text{ and }\left(1\pm 7\sqrt{2}\cos {135}^{\circ },-2\pm 7\sqrt{2}\sin {135}^{\circ }\right)\equiv (8,-9)\text{ and }(-6,5)$