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

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(B) The equation of 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 of the circle i

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$(8,5)$

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(B) The equation of 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 of the circle is $(h, k) = (1, -2)$ and the radius is $r = \sqrt{98} = 7\sqrt{2}$. For a square inscribed in a circle with sides parallel to the coordinate axes,the vertices are given by $(h \pm r\cos(\pi/4), k \pm r\sin(\pi/4))$. Since $\cos(\pi/4) = \sin(\pi/4) = 1/\sqrt{2}$,the vertices are $(1 \pm (7\sqrt{2})(1/\sqrt{2}), -2 \pm (7\sqrt{2})(1/\sqrt{2}))$. This simplifies to $(1 \pm 7, -2 \pm 7)$. The possible vertices are $(1+7, -2+7) = (8, 5)$,$(1+7, -2-7) = (8, -9)$,$(1-7, -2+7) = (-6, 5)$,and $(1-7, -2-7) = (-6, -9)$. Comparing these with the given options,$(8, 5)$ is a valid vertex.

$A$ square is inscribed in the circle $x^2+y^2-2x+4y-93=0$ with its sides parallel to the coordinate axes. Which of the following can be one of the vertices of the square?

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