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) and (8, 5)
• B. (-6, 9), (-6, -5), (8, -9) and (8, 5)
• C. (-6, -9), (-6, 5), (8, 9) and (8, 5)
• D. (-6, -9), (-6, 5), (8, -9) and (8, -5)

Answer 1

The correct option is A

(-6, -9), (-6, 5), (8, -9) and (8, 5)

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., $2\sqrt{1+4+93}$. Let l be the length of each side of the square.

Then $2l^2=(Diagonal{)}^2\Rightarrow 2l^2=(2.\sqrt{1+4+93}{)}^2$ $\Rightarrow$ l = 14

Therefore each side of the square is at a distance 7 from the centre (1, -2) of the given circle. This implies

that a = -6, b = 8, c = -9, d = 5.

Therefore the vertices of the square are (-6, -9), (-6, 5), (8, -9), (8, 5).