Thank you for reporting, we will resolve it shortly
Q.
A square is inscribed in the circle $x^2 + y^2 - 2x + 4y - 93 = 0$ with its sides parallel to the coordinate axes. The coordinates of its vertices are
Conic Sections
Solution:
Let $x = a, x = b, y = c$, and $y = d$ be the sides of the square.
The length of each diagonal of the square is equal to the diameter of the circle, i.e., $2\sqrt{98}$.
Let $l$ be the length of each side of the square. Then,
$2l^2 =$ (Diagonal)$^2$ or $l = 14$
Therefore, each side of the square is at a distance $7$ from the center $(1 ,-2)$ of the given circle.
This implies that $a = - 6, b = 8, c = -9$, and $d = 5$.
Hence, the vertices of the square are $(-6 , -9 ), (-6 , 5), (8, -9 )$, and $(8, 5)$.