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

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(C) The equation of the circle is $x^2 + y^2 - 6x + 8y - 103 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3$,$f = 4$,and $c = -103

Vedclass · Accepted Answer

$\sqrt{41}$

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(C) The equation of the circle is $x^2 + y^2 - 6x + 8y - 103 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3$,$f = 4$,and $c = -103$.The centre of the circle is $(-g, -f) = (3, -4)$.The radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + 4^2 - (-103)} = \sqrt{9 + 16 + 103} = \sqrt{128} = 8\sqrt{2}$.Since the sides of the square are parallel to the coordinate axes,the vertices are at a distance $r/\sqrt{2} = 8$ from the centre along the horizontal and vertical directions.The vertices are $(3 \pm 8, -4 \pm 8)$,which are $(11, 4), (11, -12), (-5, 4), (-5, -12)$.The distances of these vertices from the origin $(0, 0)$ are:$d_1 = \sqrt{11^2 + 4^2} = \sqrt{121 + 16} = \sqrt{137}$$d_2 = \sqrt{11^2 + (-12)^2} = \sqrt{121 + 144} = \sqrt{265}$$d_3 = \sqrt{(-5)^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}$$d_4 = \sqrt{(-5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$The minimum distance is $\sqrt{41}$.

$A$ square is inscribed in the circle $x^2 + y^2 - 6x + 8y - 103 = 0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of the square which is nearest to the origin is

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