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A square is inscribed in the circle \mathrm{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

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


Option: 3

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


Option: 4

none of these


Answers (1)

best_answer

Two diagonals will be inclined at 45^{\circ}with the axes.
Parametric equation
\mathrm{\frac{x+1}{\frac{1}{\sqrt{2}}}=\frac{y+2}{\frac{1}{\sqrt{2}}}=r}

Point \mathrm{(2,-1)(0,-3)}
Parametric equation other diagonal

\mathrm{\frac{x+1}{-\frac{1}{\sqrt{2}}}=\frac{y+2}{\frac{1}{\sqrt{2}}}=r}
Hence points are \mathrm{(0,-1),(2,-3)}

Hence (D) is the correct answer.

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Gunjita

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