A square is inscribed in the circle $x^{2} + y^{2} - 2 x + 4 y + 3 = 0,$ whose sides are parallel to the coordinate axes. One vertex of the square is

(1 + \sqrt{2}, - 2)

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(1 - \sqrt{2}, - 2)

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(1 - 2 + \sqrt{2})

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None of these

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Solution

The correct option is D None of these
We can see that the centre of the given circle is (1, -2).
Since the sides of square are parallel to the axes, therefore, first three alternates cannot be vertices of square because in first two (a and b) y = -2 and in (c) (x = 1), which passes through centre (1, -2) but it is not possible.
Therefore none of the above three is correct.

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