Question

A square is inscribed in the circle x² + y² - 6x + 8y - 103 = 0 with its sides parallel to the corrdinate axes. Then the distance of the vertex of this square which is nearest to the origin is :

• A. 13
• B. √137
• C. 6
• D. √41

Answer 1

Answer: (D)

√41

Answer 2

The equation of the circle is $x^2 + y^2 - 6x + 8y - 103 = 0$. Completing the square, we rewrite it as $(x-3)^2 + (y+4)^2 = 103 + 9 + 16 = 128$. The center of the circle is $C(3, -4)$ and the radius is $r = \sqrt{128} = 8\sqrt{2}$. Since the square is inscribed with sides parallel to the axes, its center is also $C(3, -4)$. Let the vertices of the square be $(3 \pm a, -4 \pm a)$. The distance from the center $C$ to any vertex is equal to the radius $r$. $\sqrt{((3+a)-3)^2 + ((-4+a)-(-4))^2} = 8\sqrt{2}$ $\sqrt{a^2 + a^2} = 8\sqrt{2} \implies a\sqrt{2} = 8\sqrt{2} \implies a=8$. The vertices of the square are: $(3+8, -4+8) = (11, 4)$ $(3-8, -4+8) = (-5, 4)$ $(3+8, -4-8) = (11, -12)$ $(3-8, -4-8) = (-5, -12)$ The distances of these vertices from the origin $(0,0)$ are: $\sqrt{11^2 + 4^2} = \sqrt{121+16} = \sqrt{137}$ $\sqrt{(-5)^2 + 4^2} = \sqrt{25+16} = \sqrt{41}$ $\sqrt{11^2 + (-12)^2} = \sqrt{121+144} = \sqrt{265}$ $\sqrt{(-5)^2 + (-12)^2} = \sqrt{25+144} = \sqrt{169} = 13$ The minimum distance is $\sqrt{41}$.