A square is inscribed in the circle x^{2} + y | vedclass

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(C) The given circle is $x^{2} + y^{2} - 2x + 4y - 93 = 0$.Completing the square,we get $(x - 1)^{2} + (y + 2)^{2} = 93 + 1 + 4 = 98$.The center is $(

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

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(C) The given circle is $x^{2} + y^{2} - 2x + 4y - 93 = 0$.Completing the square,we get $(x - 1)^{2} + (y + 2)^{2} = 93 + 1 + 4 = 98$.The center is $(1, -2)$ and the radius $R = \sqrt{98} = 7\sqrt{2}$.Let the side length of the square be $s$. Since the sides are parallel to the axes,the distance from the center $(1, -2)$ to each side is $s/2$.The distance from the center to a vertex is the radius $R = 7\sqrt{2}$.In a square with sides parallel to the axes,the distance from the center to a vertex is $\sqrt{(s/2)^{2} + (s/2)^{2}} = \frac{s}{\sqrt{2}}$.Equating this to the radius: $\frac{s}{\sqrt{2}} = 7\sqrt{2} \Rightarrow s = 14$.The sides are at distance $s/2 = 7$ from the center $(1, -2)$.The vertical sides are $x = 1 \pm 7$,i.e.,$x = -6$ and $x = 8$.The horizontal sides are $y = -2 \pm 7$,i.e.,$y = -9$ and $y = 5$.Thus,the vertices are $(-6, -9), (-6, 5), (8, -9), (8, 5)$.

$A$ square is inscribed in the circle $x^{2} + y^{2} - 2x + 4y - 93 = 0$ such that its sides are parallel to the coordinate axes. The coordinates of its vertices are:

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