A square is inscribed in the circle x^2+y^2-2x+4y-93=0 with its sides parallel to the co-ordinate axes. Find the co-ordinates of its vertices.

Hello Student

(x-1)^2+(y+2)^2=93+1+4=98=(7*sqrt(2))
The vertices are at a distance of 7 units (Half the side) measured in
parallel with x & y axes, from the center (1, -2) of the circle.
Hence the vertex A is (1-7, -2-7) = (-6, -9)
Vertex B is (1+7, -2-7) = (8, -9)
Vertex C is (1+7, -2+7) = (8, 5)
Vertex D is (1-7, -2+7) = (-6, 5)

Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty