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

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(A) The given circle is $x^2 + y^2 - 2x + 4y - 93 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = 2, c = -93$.The centre is $(

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

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(A) The given circle is $x^2 + y^2 - 2x + 4y - 93 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = 2, c = -93$.The centre is $(-g, -f) = (1, -2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{1 + 4 + 93} = \sqrt{98} = 7\sqrt{2}$.The diameter of the circle is $2r = 14\sqrt{2}$.Since the square is inscribed in the circle,its diagonal is equal to the diameter of the circle.Let $l$ be the side length of the square. Then $l\sqrt{2} = 14\sqrt{2}$,which gives $l = 14$.Since the sides are parallel to the coordinate axes,the distance of each side from the centre $(1, -2)$ is $l/2 = 7$.The $x$-coordinates of the sides are $1 \pm 7$,i.e.,$x = -6$ and $x = 8$.The $y$-coordinates of the sides are $-2 \pm 7$,i.e.,$y = -9$ and $y = 5$.Thus,the vertices of the square are $(-6, -9), (-6, 5), (8, -9), (8, 5)$.

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

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