Question

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

• A. $(-6,-9),(-6,5),(8,-9);(8,5)$
• B. $(-6,9),(-6,-5),(8,-9),(8,5)$
• C. $(-6,-9),(-6,5),(8,9),(8,5)$
• D. $(-6,-9),(-6,5),(8,-9),(8,-5)$

Answer 1

Answer: (A)

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 $(8,5)$.