A square is inscribed in a circle x ² + y ² + 2x + 2y + 1 = 0 and its sides are parallel to coordinate axes. Which one of the following is a vertex of the square?

Calculation:

Given,

The circle’s equation is

\(x^{2} + y^{2} + 2x + 2y + 1 = 0\)

Rewrite by completing squares:

\((x + 1)^{2} + (y + 1)^{2} = 1\)

∴ center = (-1, -1), radius r = 1

For a square inscribed in this circle with sides parallel to the axes, its diagonal equals the circle’s diameter = 2. If the square has side length a, then

\(a\sqrt{2} = 2 \;\Longrightarrow\; a = \sqrt{2}.\)

Each vertex lies a half‐side \(= \tfrac{a}{2} = \tfrac{1}{\sqrt{2}} \) from the center along both axes. Since the center is ( -1, -1), the four vertices are

\(\displaystyle \bigl(-1 \pm \tfrac{1}{\sqrt{2}},\; -1 \pm \tfrac{1}{\sqrt{2}}\bigr).\)

One of these vertices is

\(\bigl(-1 + \tfrac{1}{\sqrt{2}},\; -1 - \tfrac{1}{\sqrt{2}}\bigr).\)

Hence, the correct answer is Option 3