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A square is inscribed in the circle x^2+y^2-2 x+4 y+3=0. Its sides are parallel to the coordinate axes. Then one vertex of the square is

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

\mathrm{none \, \, of \, \, these}


Answers (1)

best_answer

The centre of the given circle is (1,-2). Since the sides of the square inscribed in the circle are parallel to the coordinate axes, so the x coordinate of any vertex cannot be equal to 1 and its y coordinate cannot be equal to -2 .

Hence none of the points given in (a),(b) or (c) can be the vertex of the square. Thus the correct answer is (d).

\therefore \quad (d)

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